Physics Electricity
Key Concepts & Formulas
| # | Concept | Quick Explanation |
|---|---|---|
| 1 | Ohm’s Law | V = I × R (Voltage = Current × Resistance) – works only at constant temperature & for ohmic conductors. |
| 2 | Resistance in series | R_total = R₁ + R₂ + … (current same, voltage divides) |
| 3 | Resistance in parallel | 1/R_total = 1/R₁ + 1/R₂ + … (voltage same, current divides) |
| 4 | Electric Power | P = V I = I² R = V²/R (units: watt) |
| 5 | Joule’s Law of Heating | H = I² R t (heat ∝ square of current) |
| 6 | Specific Resistance | ρ = R A / l (material property, unit Ω-m) |
| 7 | Conductance | G = 1/R (unit: siemen, S) |
10 Practice MCQs
1. A 220 V, 100 W bulb is connected to 110 V. The power consumed will be –
A) 100 W B) 75 W C) 50 W D) 25 WAnswer: D
Solution: Resistance of bulb R = V²/P = 220²/100 = 484 Ω. New power P′ = V′²/R = 110²/484 = 25 W.
Shortcut: P ∝ V²; voltage halved → power becomes (½)² = ¼ of original.
Concept tag: Power scaling with voltage.
2. Three 6 Ω resistors in parallel give total resistance –
A) 18 Ω B) 6 Ω C) 2 Ω D) 0.5 ΩAnswer: C
Solution: 1/R = 1/6 + 1/6 + 1/6 = 3/6 ⇒ R = 2 Ω.
Shortcut: n equal resistors R in parallel → R_total = R/n.
Concept tag: Parallel grouping.
3. The unit of specific resistance is –
A) Ω B) Ω-m C) Ω/m D) mho-mAnswer: B
Solution: ρ = R A / l → Ω × m² / m = Ω-m.
Shortcut: Memorise “rho-unit = ohm-metre”.
Concept tag: Specific resistance unit.
4. A current of 2 A flows through a 5 Ω resistor for 10 s. Heat produced is –
A) 50 J B) 100 J C) 200 J D) 500 JAnswer: C
Solution: H = I² R t = 2² × 5 × 10 = 200 J.
Shortcut: Square the current first (2² = 4) then multiply.
Concept tag: Joule heating.
5. Kirchhoff’s second law is based on conservation of –
A) Charge B) Energy C) Momentum D) MassAnswer: B
Solution: ΣIR = ΣEMF → work/energy balance in closed loop.
Shortcut: “K-2 = Energy”.
Concept tag: Kirchhoff’s laws.
6. A 1 kW heater works on 220 V. The current drawn is approximately –
A) 0.22 A B) 2.2 A C) 4.5 A D) 22 AAnswer: C
Solution: I = P/V = 1000/220 ≈ 4.5 A.
Shortcut: 1000 ÷ 220 ≈ 100 ÷ 22 ≈ 4.5.
Concept tag: Current from power.
7. The resistance of a wire is 10 Ω. If its length is doubled and area halved, the new resistance is –
A) 5 Ω B) 10 Ω C) 20 Ω D) 40 ΩAnswer: D
Solution: R ∝ l/A; l → 2l, A → A/2 ⇒ R_new = 10 × (2/0.5) = 10 × 4 = 40 Ω.
Shortcut: “double length & half area → 4× resistance”.
Concept tag: Resistance geometry.
8. The colour code of a 47 kΩ ±10% resistor is –
A) Yellow-Violet-Orange-Silver B) Yellow-Violet-Yellow-Silver C) Yellow-Violet-Red-Silver D) Yellow-Violet-Black-GoldAnswer: B
Solution: 47 × 10³ Ω → 4-7-4 (yellow-violet-yellow) & silver = ±10%.
Shortcut: “4-7-0000” → 4,7,4 bands.
Concept tag: Resistor colour code.
9. A cell of emf 1.5 V supplies 0.3 A through an external 4 Ω resistance. Its internal resistance is –
A) 0.5 Ω B) 1 Ω C) 1.5 Ω D) 2 ΩAnswer: B
Solution: V = I(R+r) ⇒ 1.5 = 0.3(4+r) ⇒ 5 = 4+r ⇒ r = 1 Ω.
Shortcut: 0.3 A × 4 Ω = 1.2 V drop across R; remaining 0.3 V across r → r = 0.3 V/0.3 A = 1 Ω.
Concept tag: Internal resistance.
10. Which material has negative temperature coefficient of resistance?
A) Copper B) Nichrome C) Germanium D) TungstenAnswer: C
Solution: Semiconductors (Ge) show resistance ↓ with temperature ↑.
Shortcut: “NTC = Semiconductor”.
Concept tag: Temperature coefficient.
5 Previous Year Questions
1. Two 2 Ω resistors in series and two such combinations in parallel. Effective resistance is – [RRB NTPC 2021]
A) 1 Ω B) 2 Ω C) 4 Ω D) 8 ΩAnswer: B
Solution: Each series branch = 2+2 = 4 Ω; two 4 Ω in parallel → 2 Ω.
Shortcut: Symmetry → average value.
Concept tag: Series-parallel combo.
2. Energy consumed by a 60 W bulb in 2 minutes is – [RRB Group-D 2019]
A) 120 J B) 720 J C) 7200 J D) 1200 JAnswer: C
Solution: E = P t = 60 W × 120 s = 7200 J.
Shortcut: 60 × 120 = 7200.
Concept tag: Energy = P × t.
3. A 200 V, 5 A motor lifts a load in 20 s. Work done is – [RRB JE 2015]
A) 20 kJ B) 2 kJ C) 10 kJ D) 200 kJAnswer: A
Solution: W = V I t = 200 × 5 × 20 = 20 000 J = 20 kJ.
Shortcut: 200 × 5 = 1000; ×20 = 20 kJ.
Concept tag: Electrical work.
4. A fuse wire has – [RRB NTPC 2016]
A) High resistance, high melting point B) Low resistance, low melting point C) High resistance, low melting point D) Low resistance, high melting pointAnswer: C
Solution: It must melt quickly → low mp; limit current → high resistance.
Shortcut: “Fuse = High R, Low mp”.
Concept tag: Fuse properties.
5. The commercial unit of electrical energy is – [RRB ALP 2018]
A) Watt B) Joule C) Kilowatt-hour D) KilowattAnswer: C
Solution: 1 kWh = 3.6 × 10⁶ J.
Shortcut: “Bill unit = kWh”.
Concept tag: Energy units.
Speed Tricks & Shortcuts
| Situation | Shortcut | Example |
|---|---|---|
| Equal resistors in parallel | R_total = R / n | 4 × 10 Ω in parallel → 2.5 Ω |
| Power at different voltage | P_new = P_old × (V_new/V_old)² | 100 W bulb on 110 V → 25 W |
| Quick heat in calories | H(cal) ≈ 0.24 I² R t | 2 A, 5 Ω, 10 s → 0.24×200 ≈ 48 cal |
| Internal resistance | r = (E – V)/I | E = 1.5 V, V = 1.2 V, I = 0.3 A → r = 1 Ω |
| Colour-code 3rd band | Add zeroes equal to band value | Yellow (4) → 0000 → 10⁴ |
Common Mistakes to Avoid
| Mistake | Why Students Make It | Correct Approach |
|---|---|---|
| Forgetting to square current in Joule’s law | Confuse P = VI with H = I²Rt | Always check which formula is asked; heat uses I² |
| Using wrong formula for parallel bulbs | Assume powers add directly | Powers add only when voltages are same; recalculate current first |
| Ignoring internal resistance in cells | Treat terminal voltage = emf | Subtract Ir drop: V = E – Ir |
| Mixing kWh with joules | kWh is 3.6 × 10⁶ J, not 1000 J | Convert to joules only when asked in SI units |
Quick Revision Flashcards
| Front | Back |
|---|---|
| Ohm’s Law statement | V = IR at constant temperature |
| SI unit of resistivity | Ohm-metre (Ω-m) |
| Conductance formula | G = 1/R (siemen) |
| Power in terms of V & R | P = V²/R |
| Energy in kWh to joules | 1 kWh = 3.6 × 10⁶ J |
| Fuse wire property | High R, low melting point |
| Kirchhoff’s 1st law | ΣI_in = ΣI_out (junction rule) |
| Kirchhoff’s 2nd law | ΣIR = ΣEMF (loop rule) |
| Temperature coefficient of semiconductor | Negative |
| Commercial unit of energy | Kilowatt-hour |
BETA
