🟡 Senior's Tip: Electricity is the BIGGEST chapter for AMU — 3-4 questions every exam. Ohm's law, series/parallel resistance, Joule's heating, and electric power are non-negotiable. If you master the 5 key formulas (I=Q/t, V=W/Q, V=IR, H=I²Rt, P=VI) and their rearrangements, you're looking at full marks. This chapter also builds the foundation for Magnetic Effects. Let's go deep!
1) Electric Charge — The Foundation
Electric charge is the fundamental property of matter responsible for electrical phenomena. It's what makes protons and electrons interact. Before we talk about current, we MUST understand charge — because current is just charge in motion!
Key facts about charge:
• Like charges repel, unlike charges attract
• Unit: coulomb (C)
• Charge on 1 electron: e = 1.6 × 10⁻¹⁹ C
• Charge on 1 proton: +1.6 × 10⁻¹⁹ C
• Charge is quantized: Q = ne (where n = integer, e = elementary charge)
• Charge is conserved (can't be created or destroyed, only transferred)
• Like charges repel, unlike charges attract
• Unit: coulomb (C)
• Charge on 1 electron: e = 1.6 × 10⁻¹⁹ C
• Charge on 1 proton: +1.6 × 10⁻¹⁹ C
• Charge is quantized: Q = ne (where n = integer, e = elementary charge)
• Charge is conserved (can't be created or destroyed, only transferred)
🟡 Senior's Tip: Quantization means charge always comes in integer multiples of e. You can never have 2.5 × 10⁻¹⁹ C of charge — it's always 1.6 × 10⁻¹⁹ × integer. This is a common trick in MCQs!
🔥 AMU PYQ: Which of the following is NOT a possible value of charge?
(A) 1.6 × 10⁻¹⁹ C (B) 3.2 × 10⁻¹⁹ C (C) 4.8 × 10⁻¹⁹ C (D) 2.0 × 10⁻¹⁹ C
Ans: D (2.0 × 10⁻¹⁹ C is not an integer multiple of 1.6 × 10⁻¹⁹; 2.0/1.6 = 1.25, not an integer)
(A) 1.6 × 10⁻¹⁹ C (B) 3.2 × 10⁻¹⁹ C (C) 4.8 × 10⁻¹⁹ C (D) 2.0 × 10⁻¹⁹ C
Ans: D (2.0 × 10⁻¹⁹ C is not an integer multiple of 1.6 × 10⁻¹⁹; 2.0/1.6 = 1.25, not an integer)
Numerical: How many electrons make up 1 C of charge?
Solution:
Q = ne → n = Q/e = 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons
So 6.25 × 10¹⁸ electrons = 1 coulomb of charge
Solution:
Q = ne → n = Q/e = 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons
So 6.25 × 10¹⁸ electrons = 1 coulomb of charge
2) Electric Current
Electric current is the rate of flow of electric charge through a conductor. If 1 C of charge passes through a point in 1 second, the current is 1 ampere. Simple!
I = Q / t
I = current (ampere, A)
Q = charge (coulomb, C)
t = time (second, s)
1 A = 1 C/s
🟡 Direction of current — DON'T GET CONFUSED:
• Conventional current: flows from positive (+) to negative (−) terminal (this is what we use in all circuit analysis)
• Electron flow: opposite direction (negative to positive) — this is what actually happens physically
• Rule: In all diagrams and calculations, mark conventional current direction (+ to −)
• Electric current is measured by an ammeter (connected in series)
• Conventional current: flows from positive (+) to negative (−) terminal (this is what we use in all circuit analysis)
• Electron flow: opposite direction (negative to positive) — this is what actually happens physically
• Rule: In all diagrams and calculations, mark conventional current direction (+ to −)
• Electric current is measured by an ammeter (connected in series)
Numerical 1: A current of 2 A flows through a wire. How much charge passes through it in 30 seconds?
Solution: Q = I × t = 2 × 30 = 60 C
Solution: Q = I × t = 2 × 30 = 60 C
Numerical 2: If 120 C of charge flows through a conductor every minute, find the current.
Solution: t = 1 min = 60 s, Q = 120 C
I = Q/t = 120/60 = 2 A
Solution: t = 1 min = 60 s, Q = 120 C
I = Q/t = 120/60 = 2 A
🔥 AMU PYQ: SI unit of electric current is:
(A) Coulomb (B) Volt (C) Ampere (D) Ohm
Ans: C (Ampere, named after André-Marie Ampère)
(A) Coulomb (B) Volt (C) Ampere (D) Ohm
Ans: C (Ampere, named after André-Marie Ampère)
🔥 AMU PYQ: One ampere is equivalent to:
(A) 1 J/s (B) 1 V/A (C) 1 C/s (D) 1 Ω/s
Ans: C (1 A = 1 C/s)
(A) 1 J/s (B) 1 V/A (C) 1 C/s (D) 1 Ω/s
Ans: C (1 A = 1 C/s)
3) Potential Difference
For current to flow, there must be a potential difference (voltage) across the conductor. Think of it as the "push" that makes charges move. A cell or battery provides this push by doing work to separate charges.
V = W / Q
V = potential difference (volt, V)
W = work done (joule, J)
Q = charge (coulomb, C)
1 V = 1 J/C
🟡 Senior's Tip to remember:
• Volt = Joule per Coulomb (energy per unit charge)
• Ammeter → series (current must flow through it)
• Voltmeter → parallel (measures voltage across a component)
• Think of voltage as the "water pressure" and current as the "flow rate" in a pipe
• Volt = Joule per Coulomb (energy per unit charge)
• Ammeter → series (current must flow through it)
• Voltmeter → parallel (measures voltage across a component)
• Think of voltage as the "water pressure" and current as the "flow rate" in a pipe
Numerical: If 50 J of work is done to move 10 C of charge between two points, what is the potential difference?
Solution: V = W/Q = 50/10 = 5 V
Solution: V = W/Q = 50/10 = 5 V
Numerical: How much work is done in moving 5 C of charge through a potential difference of 12 V?
Solution: W = V × Q = 12 × 5 = 60 J
Solution: W = V × Q = 12 × 5 = 60 J
🔥 AMU PYQ: 1 volt is equal to:
(A) 1 J/C (B) 1 C/J (C) 1 J/s (D) 1 A/s
Ans: A (1 V = 1 J/C — work done per unit charge)
(A) 1 J/C (B) 1 C/J (C) 1 J/s (D) 1 A/s
Ans: A (1 V = 1 J/C — work done per unit charge)
4) Ohm's Law ⭐ — THE MOST IMPORTANT LAW
Ohm's law is the relationship between voltage, current, and resistance. It is the single most tested concept in AMU from this chapter. Memorise it like your name!
Statement: At constant temperature, the current (I) flowing through a conductor is directly proportional to the potential difference (V) across its ends.
V ∝ I → V = I × R
V = potential difference (V)
I = current (A)
R = resistance (ohm, Ω)
Rearranged forms:
I = V / R
R = V / I
Experimental Verification of Ohm's Law:
Set up a circuit with a cell, ammeter (series), voltmeter (parallel across the resistor), rheostat (to vary current), and a key. Vary the current using the rheostat and note the voltmeter and ammeter readings. Plot a graph of V vs I.
🟡 V-I Graph for Ohmic Conductors:
• The graph is a straight line passing through the origin
• Slope of the line = resistance (R)
• Ohmic conductors: metals, copper wire, nichrome wire (obey Ohm's law)
• Non-ohmic conductors: diode, LED, filament bulb, semiconductor (do NOT obey Ohm's law)
• For non-ohmic devices, the V-I graph is NOT a straight line (curved)
• The graph is a straight line passing through the origin
• Slope of the line = resistance (R)
• Ohmic conductors: metals, copper wire, nichrome wire (obey Ohm's law)
• Non-ohmic conductors: diode, LED, filament bulb, semiconductor (do NOT obey Ohm's law)
• For non-ohmic devices, the V-I graph is NOT a straight line (curved)
Numerical 1: A potential difference of 12 V across a resistor produces a current of 3 A. Find the resistance.
Solution: R = V/I = 12/3 = 4 Ω
Solution: R = V/I = 12/3 = 4 Ω
Numerical 2: A 6 Ω resistor has a current of 0.5 A flowing through it. Find the voltage across it.
Solution: V = IR = 0.5 × 6 = 3 V
Solution: V = IR = 0.5 × 6 = 3 V
Numerical 3: A bulb draws 0.25 A current when connected to a 10 V supply. Find its resistance.
Solution: R = V/I = 10/0.25 = 40 Ω
Solution: R = V/I = 10/0.25 = 40 Ω
🔥 AMU PYQ: Ohm's law states that:
(A) V ∝ R (B) V ∝ I (C) I ∝ R (D) V ∝ 1/I
Ans: B (V ∝ I at constant temperature)
(A) V ∝ R (B) V ∝ I (C) I ∝ R (D) V ∝ 1/I
Ans: B (V ∝ I at constant temperature)
🔥 AMU PYQ: Which of the following is a non-ohmic conductor?
(A) Copper wire (B) Nichrome wire (C) Silicon diode (D) Aluminum wire
Ans: C (diode does not obey Ohm's law — it allows current in only one direction)
(A) Copper wire (B) Nichrome wire (C) Silicon diode (D) Aluminum wire
Ans: C (diode does not obey Ohm's law — it allows current in only one direction)
5) Resistance and Resistivity ⭐
Resistance is the opposition that a conductor offers to the flow of current. Think of it as friction for electric current — more resistance = less current for the same voltage.
R = ρ × L / A
R = resistance (Ω)
ρ = resistivity (Ωm) — depends only on material and temperature
L = length of conductor (m)
A = cross-sectional area (m²)
Factors affecting resistance — in detail:
- Length (L): R ∝ L — longer wire = more collisions = more resistance. Double the length → double the resistance
- Cross-sectional Area (A): R ∝ 1/A — thicker wire = more space for electrons = less resistance. Double the area → half the resistance
- Temperature: For metals, R increases with temperature (atoms vibrate more → more collisions). For semiconductors (like silicon, germanium), R decreases with temperature
- Material (ρ): Different materials have different resistivities. Silver has the lowest resistivity (best conductor after superconductors)
🟡 Resistivity (ρ) — the material's "personality":
• SI unit: ohm-metre (Ωm)
• Conductors: LOW resistivity (Silver 1.6×10⁻⁸, Copper 1.7×10⁻⁸, Aluminum 2.7×10⁻⁸ Ωm)
• Insulators: VERY HIGH resistivity (Rubber ~10¹³, Glass ~10¹⁰, Wood ~10⁸ Ωm)
• Alloys have HIGHER resistivity than pure metals (Nichrome ~10⁻⁶ Ωm → perfect for heating elements!)
• Resistivity depends ONLY on material and temperature — not on length or area!
• SI unit: ohm-metre (Ωm)
• Conductors: LOW resistivity (Silver 1.6×10⁻⁸, Copper 1.7×10⁻⁸, Aluminum 2.7×10⁻⁸ Ωm)
• Insulators: VERY HIGH resistivity (Rubber ~10¹³, Glass ~10¹⁰, Wood ~10⁸ Ωm)
• Alloys have HIGHER resistivity than pure metals (Nichrome ~10⁻⁶ Ωm → perfect for heating elements!)
• Resistivity depends ONLY on material and temperature — not on length or area!
Temperature Dependence of Resistance — NEW TOPIC:
This is frequently asked in AMU and is often misunderstood. Let's clear it up:
- Metals: As temperature ↑, atomic vibrations ↑ → more collisions with electrons → resistance ↑ (positive temperature coefficient)
- Semiconductors: As temperature ↑, more electrons become free to conduct → resistance ↓ (negative temperature coefficient)
- Thermistor: A special resistor whose resistance changes dramatically with temperature. Used in thermometers and temperature sensors. Two types: PTC (positive temp coefficient) and NTC (negative temp coefficient)
- Superconductor: A material that has ZERO resistance below a certain critical temperature (e.g., mercury below 4.2 K). Current flows forever without loss! Used in MRI machines, particle accelerators
Numerical — The Stretching Wire Problem (VERY IMPORTANT):
Q: A wire of resistance 10 Ω is stretched to double its length. What is the new resistance?
Solution:
Step 1: When a wire is stretched, its volume remains constant. V = A₁L₁ = A₂L₂
Step 2: L₂ = 2L₁ (given). So A₂ = A₁L₁/L₂ = A₁L₁/(2L₁) = A₁/2
Step 3: R = ρL/A. R₁ = ρL₁/A₁ = 10 Ω
Step 4: R₂ = ρL₂/A₂ = ρ(2L₁)/(A₁/2) = ρL₁/A₁ × 4 = 10 × 4 = 40 Ω
Key insight: When length doubles, area halves, so R becomes 4 times!
Q: A wire of resistance 10 Ω is stretched to double its length. What is the new resistance?
Solution:
Step 1: When a wire is stretched, its volume remains constant. V = A₁L₁ = A₂L₂
Step 2: L₂ = 2L₁ (given). So A₂ = A₁L₁/L₂ = A₁L₁/(2L₁) = A₁/2
Step 3: R = ρL/A. R₁ = ρL₁/A₁ = 10 Ω
Step 4: R₂ = ρL₂/A₂ = ρ(2L₁)/(A₁/2) = ρL₁/A₁ × 4 = 10 × 4 = 40 Ω
Key insight: When length doubles, area halves, so R becomes 4 times!
🔥 AMU PYQ: The resistance of a wire depends on:
(A) Length (B) Area (C) Material (D) All of these
Ans: D (R = ρL/A, so all three factors matter)
(A) Length (B) Area (C) Material (D) All of these
Ans: D (R = ρL/A, so all three factors matter)
🔥 AMU PYQ: Resistivity of a conductor depends on:
(A) Length (B) Area (C) Temperature (D) Both A and B
Ans: C (resistivity depends on material and temperature only, not on dimensions)
(A) Length (B) Area (C) Temperature (D) Both A and B
Ans: C (resistivity depends on material and temperature only, not on dimensions)
6) Series and Parallel Combination of Resistors ⭐⭐
This is THE MOST IMPORTANT topic for numericals. At least one question in AMU comes from series-parallel resistance combinations. Let's master it thoroughly.
Series Combination:
In series, resistors are joined end-to-end. The same current flows through each resistor, but the voltage divides across them.
Series combination:
R_eq = R₁ + R₂ + R₃ + ...I = same through all resistorsV = V₁ + V₂ + V₃ + ...V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
🟡 Properties of Series Circuit:
• R_eq is ALWAYS greater than the largest individual resistance
• If one component fails (bulb blows), the entire circuit breaks — this is a DISADVANTAGE
• Used in: decorative lights (fairy lights), old Christmas lights, fuses, voltage divider circuits
• All components get the SAME current (advantage for some applications)
• R_eq is ALWAYS greater than the largest individual resistance
• If one component fails (bulb blows), the entire circuit breaks — this is a DISADVANTAGE
• Used in: decorative lights (fairy lights), old Christmas lights, fuses, voltage divider circuits
• All components get the SAME current (advantage for some applications)
Parallel Combination:
In parallel, resistors are connected across the same two points. The voltage is the same across each branch, but current divides.
Parallel combination:
Special case — two resistors:
1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + ...V = same across all branchesI = I₁ + I₂ + I₃ + ...I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃Special case — two resistors:
R_eq = (R₁ × R₂) / (R₁ + R₂)
🟡 Properties of Parallel Circuit:
• R_eq is ALWAYS less than the smallest individual resistance
• If one component fails, others continue working — this is the BIG ADVANTAGE
• Used in: household wiring — all appliances get full 220 V and work independently
• Each device can be switched on/off independently
• R_eq is ALWAYS less than the smallest individual resistance
• If one component fails, others continue working — this is the BIG ADVANTAGE
• Used in: household wiring — all appliances get full 220 V and work independently
• Each device can be switched on/off independently
| Property | Series | Parallel |
|---|---|---|
| Current | Same through all resistors | Divides among branches (I = I₁+I₂+...) |
| Voltage | Divides (V = V₁+V₂+...) | Same across all branches |
| Equivalent resistance | R_eq = R₁ + R₂ + R₃ + ... | 1/R_eq = 1/R₁ + 1/R₂ + ... |
| R_eq vs individual | R_eq > largest R | R_eq < smallest R |
| If one component fails | Entire circuit breaks | Other branches unaffected |
| Advantage | Simple, all get same current | Independent operation (household wiring) |
| Disadvantage | Single point of failure | More wiring, complex design |
| Application | Decorative lights, fuse | Home wiring, power distribution |
Numerical 1 — Series: Three resistors 2 Ω, 3 Ω, and 5 Ω are connected in series. Find the equivalent resistance and current if connected to a 10 V battery.
Solution:
R_eq = 2 + 3 + 5 = 10 Ω
I = V/R_eq = 10/10 = 1 A
Voltage across each: V₁ = 1×2 = 2V, V₂ = 1×3 = 3V, V₃ = 1×5 = 5V
Check: V₁+V₂+V₃ = 2+3+5 = 10V ✓
Solution:
R_eq = 2 + 3 + 5 = 10 Ω
I = V/R_eq = 10/10 = 1 A
Voltage across each: V₁ = 1×2 = 2V, V₂ = 1×3 = 3V, V₃ = 1×5 = 5V
Check: V₁+V₂+V₃ = 2+3+5 = 10V ✓
Numerical 2 — Parallel: Three resistors 2 Ω, 3 Ω, and 6 Ω are connected in parallel. Find the equivalent resistance.
Solution:
1/R_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1
R_eq = 1 Ω
Check: R_eq (1 Ω) is less than smallest R (2 Ω) ✓
Solution:
1/R_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1
R_eq = 1 Ω
Check: R_eq (1 Ω) is less than smallest R (2 Ω) ✓
Numerical 3 — Two resistors in parallel (shortcut): A 4 Ω and 6 Ω resistor are in parallel. Find R_eq.
Solution:
R_eq = (4×6)/(4+6) = 24/10 = 2.4 Ω
Solution:
R_eq = (4×6)/(4+6) = 24/10 = 2.4 Ω
Numerical 4 — Mixed combination (IMPORTANT): Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected as: 2 Ω and 3 Ω in parallel, and this combination in series with 6 Ω. Find total resistance and current from a 12 V battery.
Solution:
Step 1: Find parallel combination of 2 Ω and 3 Ω: R_parallel = (2×3)/(2+3) = 6/5 = 1.2 Ω
Step 2: This 1.2 Ω is in series with 6 Ω: R_total = 1.2 + 6 = 7.2 Ω
Step 3: Total current: I = V/R_total = 12/7.2 = 1.67 A
Step 4: Voltage across parallel combo: V_p = I × R_p = 1.67 × 1.2 = 2 V
Step 5: Current through 2 Ω: I₂ = V_p/2 = 2/2 = 1 A
Step 6: Current through 3 Ω: I₃ = V_p/3 = 2/3 = 0.67 A
Check: I₂ + I₃ = 1 + 0.67 = 1.67 A = total current ✓
Solution:
Step 1: Find parallel combination of 2 Ω and 3 Ω: R_parallel = (2×3)/(2+3) = 6/5 = 1.2 Ω
Step 2: This 1.2 Ω is in series with 6 Ω: R_total = 1.2 + 6 = 7.2 Ω
Step 3: Total current: I = V/R_total = 12/7.2 = 1.67 A
Step 4: Voltage across parallel combo: V_p = I × R_p = 1.67 × 1.2 = 2 V
Step 5: Current through 2 Ω: I₂ = V_p/2 = 2/2 = 1 A
Step 6: Current through 3 Ω: I₃ = V_p/3 = 2/3 = 0.67 A
Check: I₂ + I₃ = 1 + 0.67 = 1.67 A = total current ✓
🔥 AMU PYQ: Two resistors 4 Ω and 6 Ω are in parallel. Equivalent resistance is:
Ans: R_eq = (4×6)/(4+6) = 24/10 = 2.4 Ω
Ans: R_eq = (4×6)/(4+6) = 24/10 = 2.4 Ω
🔥 AMU PYQ: In a series circuit, the current is:
(A) Different in each resistor (B) Same in each resistor (C) Zero (D) Maximum in the largest resistor
Ans: B (same current throughout a series circuit)
(A) Different in each resistor (B) Same in each resistor (C) Zero (D) Maximum in the largest resistor
Ans: B (same current throughout a series circuit)
🔥 AMU PYQ: Three identical resistors in parallel give 3 Ω. Their resistance in series will be:
Ans: Let each = R. For parallel: R/3 = 3 → R = 9 Ω. In series: 9+9+9 = 27 Ω
Ans: Let each = R. For parallel: R/3 = 3 → R = 9 Ω. In series: 9+9+9 = 27 Ω
7) Heating Effect of Electric Current — Joule's Law
When current flows through a resistor, electrical energy is converted into heat energy. This is the heating effect of current (also called Joule heating, I²R loss).
Joule's Law of Heating:
Also written as:
H = I² × R × t
H = heat produced (joules, J)
I = current (A)
R = resistance (Ω)
t = time (s)
Also written as:
H = V × I × t (using V = IR)H = (V² / R) × t
🟡 Which formula to use?
• Use H = I²Rt when R is constant and I is given
• Use H = VIt when V and I are given
• Use H = V²t/R when V is constant (like household circuits)
• All three are equivalent — just different forms!
• Use H = I²Rt when R is constant and I is given
• Use H = VIt when V and I are given
• Use H = V²t/R when V is constant (like household circuits)
• All three are equivalent — just different forms!
Applications of the Heating Effect — Real World:
- Electric Heater / Iron / Toaster / Kettle: Use nichrome element — high resistivity (generates lots of heat) and high melting point (doesn't melt). Nichrome is an alloy of Ni, Cr, Fe
- Electric Bulb: Filament heats to ~3000°C and glows white-hot → produces light. Filament is tungsten (melting point ~3400°C, very high). The bulb is filled with inert gas (Ar/N₂) to prevent oxidation
- Electric Fuse: Safety device. Thin wire of low melting point (tin-lead alloy). If current exceeds safe limit, fuse melts → circuit breaks → protects appliances
- Electric Geyser / Immersion Rod: Coil of heating element immersed in water. Heat transfers to water by conduction
Numerical 1: A 5 Ω resistor has a current of 2 A flowing through it for 10 minutes. Find the heat produced.
Solution:
t = 10 min = 10 × 60 = 600 s
H = I²Rt = 2² × 5 × 600 = 4 × 5 × 600 = 12,000 J = 12 kJ
Solution:
t = 10 min = 10 × 60 = 600 s
H = I²Rt = 2² × 5 × 600 = 4 × 5 × 600 = 12,000 J = 12 kJ
Numerical 2: A bulb draws 0.5 A at 220 V. Find the heat produced in 1 hour.
Solution:
t = 1 hour = 3600 s
H = VIt = 220 × 0.5 × 3600 = 396,000 J = 396 kJ
Solution:
t = 1 hour = 3600 s
H = VIt = 220 × 0.5 × 3600 = 396,000 J = 396 kJ
Numerical 3: An electric iron of resistance 50 Ω operates at 220 V for 30 minutes. Find heat produced.
Solution:
t = 30 × 60 = 1800 s
H = V²t/R = (220² × 1800)/50 = (48400 × 1800)/50 = 87120000/50 = 1,742,400 J ≈ 1.74 MJ
Solution:
t = 30 × 60 = 1800 s
H = V²t/R = (220² × 1800)/50 = (48400 × 1800)/50 = 87120000/50 = 1,742,400 J ≈ 1.74 MJ
🔥 AMU PYQ: Heat produced in a resistor is given by:
(A) H = IRt (B) H = I²Rt (C) H = I²R/t (D) H = VIt
Ans: B (Joule's law: H = I²Rt)
(A) H = IRt (B) H = I²Rt (C) H = I²R/t (D) H = VIt
Ans: B (Joule's law: H = I²Rt)
🔥 AMU PYQ: Which material is used for the filament of an electric bulb?
(A) Copper (B) Nichrome (C) Tungsten (D) Aluminium
Ans: C (tungsten — high melting point ~3400°C)
(A) Copper (B) Nichrome (C) Tungsten (D) Aluminium
Ans: C (tungsten — high melting point ~3400°C)
8) Electric Power ⭐⭐
Electric power is the rate at which electrical energy is consumed (or produced). Every appliance has a power rating — e.g., a "100 W bulb" means it consumes 100 J of electrical energy per second.
Three forms of Power (memorise all!):
P = V × I (most common — using P = VI)P = I² × R (when I and R known)P = V² / R (when V and R known — useful for household circuits)P = power (watt, W)
1 W = 1 J/s
1 kW = 1000 W
1 hp (horsepower) = 746 W
🟡 Senior's Tip — Power Formula Selection:
• P = VI → Use when V and I are given (straightforward)
• P = I²R → Use when resistors are in series (I same)
• P = V²/R → Use when resistors are in parallel (V same)
• Why? In series, I is constant, so P ∝ R (higher R = more power). In parallel, V is constant, so P ∝ 1/R (higher R = less power). This is a COMMON TEST TRICK!
• P = VI → Use when V and I are given (straightforward)
• P = I²R → Use when resistors are in series (I same)
• P = V²/R → Use when resistors are in parallel (V same)
• Why? In series, I is constant, so P ∝ R (higher R = more power). In parallel, V is constant, so P ∝ 1/R (higher R = less power). This is a COMMON TEST TRICK!
Commercial Unit — Kilowatt-hour (kWh):
The joule is too small for household energy calculations. So electricity companies measure energy in kilowatt-hours.
1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J
Energy (kWh) = Power (kW) × Time (h)
Example: A 100 W bulb used for 10 hours
Energy = 0.1 kW × 10 h = 1 kWh (1 unit of electricity)
Numerical — Electricity Bill: A household uses: 5 LED bulbs (10 W each) for 8 hours/day, 1 TV (120 W) for 6 hours/day, 1 fridge (150 W) for 24 hours/day. Find daily consumption in kWh and monthly bill at ₹8/kWh.
Solution:
Bulbs: Total power = 5 × 10 = 50 W = 0.05 kW. Energy = 0.05 × 8 = 0.4 kWh
TV: 120 W = 0.12 kW. Energy = 0.12 × 6 = 0.72 kWh
Fridge: 150 W = 0.15 kW. Energy = 0.15 × 24 = 3.6 kWh
Total daily = 0.4 + 0.72 + 3.6 = 4.72 kWh
Monthly (30 days) = 4.72 × 30 = 141.6 kWh
Bill = 141.6 × 8 = ₹1,132.80
Solution:
Bulbs: Total power = 5 × 10 = 50 W = 0.05 kW. Energy = 0.05 × 8 = 0.4 kWh
TV: 120 W = 0.12 kW. Energy = 0.12 × 6 = 0.72 kWh
Fridge: 150 W = 0.15 kW. Energy = 0.15 × 24 = 3.6 kWh
Total daily = 0.4 + 0.72 + 3.6 = 4.72 kWh
Monthly (30 days) = 4.72 × 30 = 141.6 kWh
Bill = 141.6 × 8 = ₹1,132.80
Numerical — Fuse Rating: A 1000 W heater operates at 220 V. What fuse rating should be used?
Solution:
I = P/V = 1000/220 = 4.55 A
Fuse rating should be slightly above operating current → use 5 A fuse
Solution:
I = P/V = 1000/220 = 4.55 A
Fuse rating should be slightly above operating current → use 5 A fuse
Numerical — Power from resistance: A 10 Ω resistor draws 2 A. Find power.
Solution:
P = I²R = 2² × 10 = 4 × 10 = 40 W
Also: V = IR = 2 × 10 = 20 V. Check: P = VI = 20 × 2 = 40 W ✓
Solution:
P = I²R = 2² × 10 = 4 × 10 = 40 W
Also: V = IR = 2 × 10 = 20 V. Check: P = VI = 20 × 2 = 40 W ✓
🔥 AMU PYQ: A 60 W bulb is used for 5 hours daily. Energy consumed in 30 days is:
Ans: Power = 60 W = 0.06 kW. Energy per day = 0.06 × 5 = 0.3 kWh
30 days → 0.3 × 30 = 9 kWh
Ans: Power = 60 W = 0.06 kW. Energy per day = 0.06 × 5 = 0.3 kWh
30 days → 0.3 × 30 = 9 kWh
🔥 AMU PYQ: 1 kWh = ?
(A) 3.6 × 10³ J (B) 3.6 × 10⁴ J (C) 3.6 × 10⁵ J (D) 3.6 × 10⁶ J
Ans: D (1 kWh = 1000 × 3600 = 3.6 × 10⁶ J)
(A) 3.6 × 10³ J (B) 3.6 × 10⁴ J (C) 3.6 × 10⁵ J (D) 3.6 × 10⁶ J
Ans: D (1 kWh = 1000 × 3600 = 3.6 × 10⁶ J)
🔥 AMU PYQ: A 500 W heater runs on 250 V. Its resistance is:
Ans: R = V²/P = 250²/500 = 62500/500 = 125 Ω
Ans: R = V²/P = 250²/500 = 62500/500 = 125 Ω
🔥 AMU PYQ: An electric bulb is rated 60 W, 220 V. The current drawn is:
Ans: I = P/V = 60/220 = 0.273 A
Ans: I = P/V = 60/220 = 0.273 A
9) Electric Cell/Battery and Internal Resistance — NEW
A cell is a device that converts chemical energy into electrical energy. When we say "1.5 V cell", that's the electromotive force (emf) — the maximum potential difference the cell can provide. But real cells have some internal resistance too!
Types of Cells:
- Dry Cell (Primary Cell): Leclanché cell. Used in torches, remotes, toys. Non-rechargeable. Zn anode, C cathode, NH₄Cl paste electrolyte. Voltage ~1.5 V. Once chemicals are used up, the cell is dead
- Secondary Cell (Rechargeable): Can be recharged by passing current through it. Examples: Lead-acid battery (car battery, ~12 V), Ni-Cd battery, Li-ion battery (phones/laptops)
🟡 Internal Resistance (r) — Concept:
Every real cell has some internal resistance (r). When current flows, some voltage drops across this internal resistance.
• When circuit is open (I = 0): V = ε (terminal voltage = emf)
• When current flows: V = ε − Ir (terminal voltage is LESS than emf)
• High internal resistance → cell heats up → poor performance
Every real cell has some internal resistance (r). When current flows, some voltage drops across this internal resistance.
Terminal voltage (V) = emf (ε) − Ir• When circuit is open (I = 0): V = ε (terminal voltage = emf)
• When current flows: V = ε − Ir (terminal voltage is LESS than emf)
• High internal resistance → cell heats up → poor performance
10) Kirchhoff's Laws (Basic, Class 11 Level) — NEW
Kirchhoff's laws are used to analyze complex circuits where Ohm's law and simple series/parallel formulas aren't enough. For AMU, you need the basic idea and simple problems.
Kirchhoff's Current Law (KCL) — Junction Rule:
Statement: The sum of currents entering a junction equals the sum of currents leaving that junction.
Σ I_in = Σ I_out
Meaning: Current is conserved. What goes in must come out.
Kirchhoff's Voltage Law (KVL) — Loop Rule:
Statement: The sum of potential drops around any closed loop equals the sum of potential rises (emfs). OR: The algebraic sum of all potential differences in a closed loop is zero.
Σ V = 0 (around a closed loop)
Meaning: Energy is conserved. Going around a loop, you end up where you started.
Simple KVL Problem: A circuit has a 12 V battery and two resistors 2 Ω and 4 Ω in series. Using KVL, find current.
Solution:
Loop: 12 V − 2I − 4I = 0
12 − 6I = 0 → I = 12/6 = 2 A
(Same as Ohm's law: R_eq = 6 Ω, I = 12/6 = 2 A ✓)
Solution:
Loop: 12 V − 2I − 4I = 0
12 − 6I = 0 → I = 12/6 = 2 A
(Same as Ohm's law: R_eq = 6 Ω, I = 12/6 = 2 A ✓)
11) Household Wiring and Safety ⭐
Understanding how electricity reaches your home is important for safety. The Indian household supply is 220 V AC at 50 Hz.
Three Wires in Household Circuits:
| Wire | Color | Purpose |
|---|---|---|
| Live (Phase) | Red / Brown | Carries current from main supply to appliance. Dangerous if touched! Connected to the switch |
| Neutral | Black / Blue | Returns current back to supply. At zero potential (earth potential) |
| Earth | Green / Yellow | Safety wire. Connected to a metal plate buried in ground. Provides path for leakage current |
Key Safety Devices:
- Fuse: Thin wire of low melting point (tin-lead alloy). Connected in series with the live wire. If current > safe value → fuse melts → circuit broken. Rated in amperes (5 A for lights/fans, 15 A for AC/geyser/heater). NEVER use a higher-rated fuse than specified!
- MCB (Miniature Circuit Breaker): Modern replacement for fuse. Uses electromagnetic effect. Automatically switches off when current exceeds limit. Can be reset (unlike fuse which needs replacement)
- Earthing: The green wire connects the metallic body of appliances to the earth. If the live wire touches the body, current flows to earth through the earth wire instead of through the user → prevents electric shock
- 3-Pin Plug: Top pin = Earth (longer, thicker), Left pin = Neutral, Right pin = Live. The earth pin is longer so it makes contact first when plugging in
Overloading and Short Circuit:
- Overloading: Too many appliances on one circuit → current exceeds rated value → wires heat up → fire risk. Also happens when the total power demand exceeds what the circuit can handle
- Short Circuit: When live and neutral wires touch directly (due to damaged insulation or accident) → very low resistance → huge current surge → sparks → fire
- Both overloading and short circuit are prevented by fuse/MCB
🟡 Senior's Safety Tips for AMU:
• Fuse is ALWAYS in series with LIVE wire (never in neutral or parallel)
• Earthing = green wire, provides safety against shock
• Household appliances are in PARALLEL (each gets 220 V independently)
• Switches are always on the LIVE wire — so when off, the appliance is completely disconnected from live supply
• Never touch electrical switches with wet hands (water reduces body resistance → more current flows through you!)
• MCB > Fuse because MCB can be reset
• Fuse is ALWAYS in series with LIVE wire (never in neutral or parallel)
• Earthing = green wire, provides safety against shock
• Household appliances are in PARALLEL (each gets 220 V independently)
• Switches are always on the LIVE wire — so when off, the appliance is completely disconnected from live supply
• Never touch electrical switches with wet hands (water reduces body resistance → more current flows through you!)
• MCB > Fuse because MCB can be reset
🔥 AMU PYQ: A fuse is connected in:
(A) Parallel (B) Series (C) Both (D) Neither
Ans: B (series with live wire)
(A) Parallel (B) Series (C) Both (D) Neither
Ans: B (series with live wire)
🔥 AMU PYQ: The purpose of earthing is to:
(A) Save electricity (B) Prevent electric shock (C) Increase current (D) Decrease voltage
Ans: B (earthing provides a safe path for leakage current to flow to the ground)
(A) Save electricity (B) Prevent electric shock (C) Increase current (D) Decrease voltage
Ans: B (earthing provides a safe path for leakage current to flow to the ground)
12) PYQ Battle — AMU Past Questions on Electricity
🔥 Q1. A current of 2 A flows for 10 s. Charge transferred is:
Ans: Q = I × t = 2 × 10 = 20 C
Ans: Q = I × t = 2 × 10 = 20 C
🔥 Q2. A wire of resistance 10 Ω is stretched to double its length. New resistance is:
Ans: Length doubles, area halves → R₂ = 4R₁ = 10 × 4 = 40 Ω
Ans: Length doubles, area halves → R₂ = 4R₁ = 10 × 4 = 40 Ω
🔥 Q3. Three resistors 2 Ω, 3 Ω, 5 Ω in series. Equivalent resistance is:
Ans: R_eq = 2 + 3 + 5 = 10 Ω
Ans: R_eq = 2 + 3 + 5 = 10 Ω
🔥 Q4. Power of a device is 100 W, voltage is 200 V. Current is:
Ans: I = P/V = 100/200 = 0.5 A
Ans: I = P/V = 100/200 = 0.5 A
🔥 Q5. Which of the following has the highest resistivity?
(A) Copper (B) Nichrome (C) Silver (D) Aluminium
Ans: B (alloys have higher resistivity than pure metals)
(A) Copper (B) Nichrome (C) Silver (D) Aluminium
Ans: B (alloys have higher resistivity than pure metals)
🔥 Q6. A 500 W heater runs on 250 V. Its resistance is:
Ans: R = V²/P = 250²/500 = 62500/500 = 125 Ω
Ans: R = V²/P = 250²/500 = 62500/500 = 125 Ω
🔥 Q7. In household circuits, appliances are connected in:
(A) Series (B) Parallel (C) Both (D) Neither
Ans: B (parallel → each gets full voltage, independent operation)
(A) Series (B) Parallel (C) Both (D) Neither
Ans: B (parallel → each gets full voltage, independent operation)
🔥 Q8. The SI unit of resistivity is:
(A) Ω (B) Ωm (C) Ω/m (D) m/Ω
Ans: B (ohm-metre)
(A) Ω (B) Ωm (C) Ω/m (D) m/Ω
Ans: B (ohm-metre)
🔥 Q9. A 2 Ω and 3 Ω resistor in parallel. Equivalent resistance is:
Ans: R_eq = (2×3)/(2+3) = 6/5 = 1.2 Ω
Ans: R_eq = (2×3)/(2+3) = 6/5 = 1.2 Ω
🔥 Q10. What is the commercial unit of electrical energy?
(A) Watt (B) Joule (C) kWh (D) Ampere
Ans: C (kilowatt-hour)
(A) Watt (B) Joule (C) kWh (D) Ampere
Ans: C (kilowatt-hour)
🔥 Q11. Which of the following does NOT obey Ohm's law?
(A) Copper wire (B) Aluminium wire (C) Diode (D) Nichrome wire
Ans: C (diode is non-ohmic)
(A) Copper wire (B) Aluminium wire (C) Diode (D) Nichrome wire
Ans: C (diode is non-ohmic)
🔥 Q12. A fuse wire is made of:
(A) Copper (B) Tungsten (C) Tin-lead alloy (D) Nichrome
Ans: C (tin-lead alloy has low melting point)
(A) Copper (B) Tungsten (C) Tin-lead alloy (D) Nichrome
Ans: C (tin-lead alloy has low melting point)
🔥 Q13. The resistance of a conductor is 5 Ω at 0°C and 6 Ω at 100°C. Its temperature coefficient of resistance is:
Ans: α = (R₂−R₁)/(R₁×ΔT) = (6−5)/(5×100) = 1/500 = 0.002 /°C
Ans: α = (R₂−R₁)/(R₁×ΔT) = (6−5)/(5×100) = 1/500 = 0.002 /°C
🔥 Q14. 1 horsepower (hp) = ?
(A) 746 W (B) 1000 W (C) 736 W (D) 550 W
Ans: A (1 hp = 746 W)
(A) 746 W (B) 1000 W (C) 736 W (D) 550 W
Ans: A (1 hp = 746 W)
🔥 Q15. Two wires of same material have lengths in ratio 2:3 and radii in ratio 3:2. Ratio of their resistances is:
Ans: R = ρL/A = ρL/πr². R₁/R₂ = (L₁/L₂) × (r₂²/r₁²) = (2/3) × (4/9) = 8/27.
Ratio = 8:27
Ans: R = ρL/A = ρL/πr². R₁/R₂ = (L₁/L₂) × (r₂²/r₁²) = (2/3) × (4/9) = 8/27.
Ratio = 8:27
🔥 Q16. The equivalent resistance of three 6 Ω resistors in parallel is:
Ans: 1/R_eq = 1/6 + 1/6 + 1/6 = 3/6 = 1/2 → R_eq = 2 Ω
Ans: 1/R_eq = 1/6 + 1/6 + 1/6 = 3/6 = 1/2 → R_eq = 2 Ω
🔥 Q17. A bulb rated 100 W, 220 V is connected to a 110 V supply. The power consumed will be:
Ans: First find R: R = V²/P = 220²/100 = 484 Ω. Then at 110 V: P = V²/R = 110²/484 = 12100/484 = 25 W (¼ of original!)
Ans: First find R: R = V²/P = 220²/100 = 484 Ω. Then at 110 V: P = V²/R = 110²/484 = 12100/484 = 25 W (¼ of original!)
🔥 Q18. An ammeter is always connected in:
(A) Series (B) Parallel (C) Both (D) Depends
Ans: A (series — to measure the current flowing through the circuit)
(A) Series (B) Parallel (C) Both (D) Depends
Ans: A (series — to measure the current flowing through the circuit)
🔥 Q19. The heating element in an electric iron is made of:
(A) Copper (B) Silver (C) Nichrome (D) Tungsten
Ans: C (nichrome — high resistivity, high melting point, does not oxidize easily)
(A) Copper (B) Silver (C) Nichrome (D) Tungsten
Ans: C (nichrome — high resistivity, high melting point, does not oxidize easily)
🔥 Q20. Kilowatt-hour is the unit of:
(A) Power (B) Current (C) Energy (D) Voltage
Ans: C (energy — commercial unit of electrical energy)
(A) Power (B) Current (C) Energy (D) Voltage
Ans: C (energy — commercial unit of electrical energy)
13) Additional Practice Problems with Step-by-Step Solutions
Practice Problem 1: A 12 V battery is connected across a combination of 3 Ω and 6 Ω in series with a 4 Ω resistor. Find total current and voltage across 3 Ω resistor.
Solution:
R_total = 3 + 6 + 4 = 13 Ω
I = V/R = 12/13 = 0.923 A
V₃ = I × 3 = 0.923 × 3 = 2.77 V
Solution:
R_total = 3 + 6 + 4 = 13 Ω
I = V/R = 12/13 = 0.923 A
V₃ = I × 3 = 0.923 × 3 = 2.77 V
Practice Problem 2: Find the equivalent resistance between A and B if two 8 Ω resistors are in parallel and this combo is in series with a 4 Ω resistor.
Solution:
R_parallel = (8×8)/(8+8) = 64/16 = 4 Ω
R_total = 4 + 4 = 8 Ω
Solution:
R_parallel = (8×8)/(8+8) = 64/16 = 4 Ω
R_total = 4 + 4 = 8 Ω
Practice Problem 3: A current of 0.5 A flows through a 200 Ω resistor for 30 minutes. Find (a) heat produced, (b) power consumed.
Solution:
(a) t = 30 × 60 = 1800 s. H = I²Rt = (0.5)² × 200 × 1800 = 0.25 × 200 × 1800 = 90,000 J = 90 kJ
(b) P = I²R = 0.25 × 200 = 50 W
Solution:
(a) t = 30 × 60 = 1800 s. H = I²Rt = (0.5)² × 200 × 1800 = 0.25 × 200 × 1800 = 90,000 J = 90 kJ
(b) P = I²R = 0.25 × 200 = 50 W
Practice Problem 4: A 2 kW electric geyser is used for 30 minutes daily. Find the energy consumed in 30 days in kWh and the cost at ₹10/kWh.
Solution:
Power = 2 kW, Time per day = 0.5 h
Energy per day = 2 × 0.5 = 1 kWh
Energy in 30 days = 1 × 30 = 30 kWh
Cost = 30 × 10 = ₹300
Solution:
Power = 2 kW, Time per day = 0.5 h
Energy per day = 2 × 0.5 = 1 kWh
Energy in 30 days = 1 × 30 = 30 kWh
Cost = 30 × 10 = ₹300
Practice Problem 5: Three cells of emf 1.5 V each and internal resistance 0.1 Ω each are connected in series to a 2 Ω external resistor. Find current in the circuit.
Solution:
Total emf = 3 × 1.5 = 4.5 V
Total internal resistance = 3 × 0.1 = 0.3 Ω
Total resistance = 0.3 + 2 = 2.3 Ω
I = ε/(R+r) = 4.5/2.3 = 1.96 A
Solution:
Total emf = 3 × 1.5 = 4.5 V
Total internal resistance = 3 × 0.1 = 0.3 Ω
Total resistance = 0.3 + 2 = 2.3 Ω
I = ε/(R+r) = 4.5/2.3 = 1.96 A
Practice Problem 6: If 0.5 A current flows through a 10 Ω resistor, find the potential difference across it and the power dissipated.
Solution:
V = IR = 0.5 × 10 = 5 V
P = VI = 5 × 0.5 = 2.5 W
Also: P = I²R = 0.25 × 10 = 2.5 W ✓
Solution:
V = IR = 0.5 × 10 = 5 V
P = VI = 5 × 0.5 = 2.5 W
Also: P = I²R = 0.25 × 10 = 2.5 W ✓
Practice Problem 7: A household circuit has a 5 A fuse. How many 60 W bulbs can be safely used at 220 V?
Solution:
Current per bulb = P/V = 60/220 = 0.273 A
Maximum bulbs = 5/0.273 = 18.3 → 18 bulbs (safety, so take integer ≤ 18)
Solution:
Current per bulb = P/V = 60/220 = 0.273 A
Maximum bulbs = 5/0.273 = 18.3 → 18 bulbs (safety, so take integer ≤ 18)
Practice Problem 8: Find the resistance of a copper wire of length 1 m and cross-sectional area 1 mm². (ρ_Cu = 1.7 × 10⁻⁸ Ωm)
Solution:
A = 1 mm² = 1 × 10⁻⁶ m²
R = ρL/A = (1.7 × 10⁻⁸ × 1) / (1 × 10⁻⁶) = 1.7 × 10⁻² = 0.017 Ω
That's why copper wires have very low resistance!
Solution:
A = 1 mm² = 1 × 10⁻⁶ m²
R = ρL/A = (1.7 × 10⁻⁸ × 1) / (1 × 10⁻⁶) = 1.7 × 10⁻² = 0.017 Ω
That's why copper wires have very low resistance!
Practice Problem 9: Two bulbs rated 60 W, 220 V and 100 W, 220 V are connected in series across 220 V. Which bulb will glow brighter?
Solution:
R₁ = V²/P₁ = 220²/60 = 806.7 Ω (60 W bulb has higher resistance)
R₂ = V²/P₂ = 220²/100 = 484 Ω
In series, P ∝ R (since I same, P = I²R)
So 60 W bulb has higher R → more power dissipation → 60 W bulb glows brighter in series!
Solution:
R₁ = V²/P₁ = 220²/60 = 806.7 Ω (60 W bulb has higher resistance)
R₂ = V²/P₂ = 220²/100 = 484 Ω
In series, P ∝ R (since I same, P = I²R)
So 60 W bulb has higher R → more power dissipation → 60 W bulb glows brighter in series!
Practice Problem 10: In the above problem, if they're connected in parallel, which glows brighter?
Solution:
In parallel, P ∝ 1/R (since V same, P = V²/R)
100 W bulb has lower R → more power → 100 W bulb glows brighter in parallel.
Key takeaway: Bulb ratings (60 W, 100 W) are for the rated voltage (220 V). The actual brightness depends on circuit connection!
Solution:
In parallel, P ∝ 1/R (since V same, P = V²/R)
100 W bulb has lower R → more power → 100 W bulb glows brighter in parallel.
Key takeaway: Bulb ratings (60 W, 100 W) are for the rated voltage (220 V). The actual brightness depends on circuit connection!
14) Common Mistakes — DON'T DO THESE!
🟡 Senior's Warning — Top 10 Mistakes Students Make:
1. ❌ Confusing series and parallel formulas: R_eq = R₁+R₂+... is for SERIES, not parallel. 1/R_eq = 1/R₁+1/R₂+... is for PARALLEL
2. ❌ Unit errors: Time in seconds for H = I²Rt, but hours for kWh. Convert min→s (×60) and hours→s (×3600) properly
3. ❌ P = VI vs P = I²R: In series, use P = I²R (since I constant). In parallel, use P = V²/R (since V constant)
4. ❌ Fuse connection: Fuse is in SERIES with LIVE wire, not neutral, not parallel
5. ❌ Ammeter vs Voltmeter: Ammeter → series. Voltmeter → parallel. Mixing these up is common!
6. ❌ Stretching wire: When length doubles, area halves → R becomes 4×, not 2× (volume conservation!)
7. ❌ Power vs Energy: Power (W, kW) is rate. Energy (J, kWh) is total consumption. Bill is for energy (kWh), not power
8. ❌ Ohmic vs Non-ohmic: Not all conductors obey Ohm's law. Diode, LED, transistor are non-ohmic
9. ❌ Conventional vs Electron flow: In all circuit problems, use conventional current (+ to −)
10. ❌ 1 kWh = 3.6 × 10⁶ J: Many students write 3.6 × 10³ or 10⁵. It's 10⁶ (million)!
1. ❌ Confusing series and parallel formulas: R_eq = R₁+R₂+... is for SERIES, not parallel. 1/R_eq = 1/R₁+1/R₂+... is for PARALLEL
2. ❌ Unit errors: Time in seconds for H = I²Rt, but hours for kWh. Convert min→s (×60) and hours→s (×3600) properly
3. ❌ P = VI vs P = I²R: In series, use P = I²R (since I constant). In parallel, use P = V²/R (since V constant)
4. ❌ Fuse connection: Fuse is in SERIES with LIVE wire, not neutral, not parallel
5. ❌ Ammeter vs Voltmeter: Ammeter → series. Voltmeter → parallel. Mixing these up is common!
6. ❌ Stretching wire: When length doubles, area halves → R becomes 4×, not 2× (volume conservation!)
7. ❌ Power vs Energy: Power (W, kW) is rate. Energy (J, kWh) is total consumption. Bill is for energy (kWh), not power
8. ❌ Ohmic vs Non-ohmic: Not all conductors obey Ohm's law. Diode, LED, transistor are non-ohmic
9. ❌ Conventional vs Electron flow: In all circuit problems, use conventional current (+ to −)
10. ❌ 1 kWh = 3.6 × 10⁶ J: Many students write 3.6 × 10³ or 10⁵. It's 10⁶ (million)!
15) Quick Revision — One-Liner Summary
| Concept | Formula / Key Point |
|---|---|
| Charge | Q = ne, e = 1.6×10⁻¹⁹ C, Like repel, unlike attract |
| Current | I = Q/t (A), Ammeter in series, Conventional: + to − |
| Potential Difference | V = W/Q (V), Voltmeter in parallel, 1 V = 1 J/C |
| Ohm's Law | V = IR (at constant temp), V-I graph = straight line through origin |
| Resistance | R = ρL/A, R ∝ L, R ∝ 1/A, depends on temp & material |
| Resistivity | ρ = RA/L (Ωm), Cu: 1.7×10⁻⁸, depends on material & temp only |
| Series | R_eq = R₁+R₂+..., I same, V divides |
| Parallel | 1/R_eq = 1/R₁+1/R₂+..., V same, I divides |
| Joule's Heating | H = I²Rt = VIt = V²t/R |
| Electric Power | P = VI = I²R = V²/R (W), 1 hp = 746 W |
| Commercial Energy | 1 kWh = 3.6×10⁶ J, Energy = P(kW) × t(h) |
| Fuse | Low melting point alloy, series with live wire, safety device |
| Household Wiring | Parallel connection, 220 V AC, separate Live/Neutral/Earth |
| Earthing | Green wire, prevents electric shock |
| Nichrome | Heating element (high ρ, high melting point) |
| Tungsten | Bulb filament (very high melting point ~3400°C) |
| KCL | Σ I_in = Σ I_out at a junction |
| KVL | Σ V = 0 around a closed loop |
🟡 Final Senior's Message:
✅ This chapter is 90% formula application. Memorise the 5 key formulas: I=Q/t, V=W/Q, V=IR, H=I²Rt, P=VI. Know when to use each form. Practice 20+ numericals and you'll crush the AMU exam.
Master these and you're guaranteed full marks in Electricity!
✅ This chapter is 90% formula application. Memorise the 5 key formulas: I=Q/t, V=W/Q, V=IR, H=I²Rt, P=VI. Know when to use each form. Practice 20+ numericals and you'll crush the AMU exam.
Master these and you're guaranteed full marks in Electricity!
✅ Electricity — DONE! Next up: Magnetic Effect of Electric Current 🧲