PA07 — Current Electricity & Circuits

✈ Physics – PA07 · AFCAT General Awareness AFCAT Level ★ High Priority

Electricity is a high-yield, high-scoring chapter for AFCAT. Ohm's law numericals, series vs parallel resistance, and power calculations appear in almost every exam. The concepts are straightforward — master the formulas and practice one example of each type.

📌 AFCAT Focus: Ohm's Law V = IR calculations; series resistance (adds up); parallel resistance (use product/sum for two resistors); which bulb glows brighter; power P = V²/R (at same voltage, lower resistance = higher power = brighter); household electricity (fuse on live wire, earthing prevents shock, India = 230V 50Hz).

1. Ohm's Law & Resistance

Core Formulae — Ohm's Law:

● Ohm's Law: V = IR    (Voltage = Current × Resistance)
    V in Volts (V); I in Amperes (A); R in Ohms (Ω)

● Resistance depends on:
    • Material (resistivity ρ): R = ρL/A
    • Length: R ∝ L (longer wire → more resistance)
    • Area: R ∝ 1/A (thicker wire → less resistance)
    • Temperature: metals → R increases with temperature
    • Semiconductors/thermistors → R decreases with temperature

● Resistivity order: Silver < Copper < Gold < Iron < Nichrome

2. Series vs Parallel Circuits

Fig. 1 — Series and Parallel Connections: Key Rules and Practical Difference

3. Power & Heating Effect (Joule's Law)

Electrical Power & Energy:

● P = VI = I²R = V²/R    [Unit: Watt (W)]
● Joule's Law of Heating: H = I²Rt    [Unit: Joule]
● Electrical energy: E = Pt    [Unit: kWh = 1 unit of electricity]
    1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J

Which bulb glows brighter at same voltage?
P = V²/R → at same V, lower resistance = more power = brighter
So: 100W bulb has LOWER resistance than a 60W bulb!

✎ Worked Example — Ohm's Law

Three resistors of 4Ω, 6Ω, and 12Ω are connected in parallel. Find the effective resistance.

1/R = 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12 = 1/2
Therefore R = 2 Ω. (Note: parallel resistance is always LESS than the smallest individual resistance — 2 < 4 ✓)

✔ R_parallel = 2 Ω

4. Household Electricity & Safety

📏 Fuse & MCB

Fuse = thin wire with low melting point
Placed on the LIVE wire (never neutral)
Melts and breaks circuit when current exceeds safe limit
Protects appliances from overload/short circuit
MCB (Miniature Circuit Breaker): reusable fuse — trips and can be reset

🔋 Earthing & AC Supply

Earthing: connects metal body of appliance to ground — prevents electric shock if current leaks to body
India mains supply: 230V AC, 50 Hz
Live wire: red/brown; Neutral: black/blue; Earth: green/yellow
All home appliances connected in parallel — each gets full 230V; independent switching

📝 AFCAT PYQs — Current Electricity

Q1. Three resistors each of 6Ω are connected in parallel. The combined resistance is: AFCAT PYQ

(a) 18 Ω(b) 6 Ω(c) 2 Ω(d) 3 Ω

✔ Answer: (c) 2 Ω

For n equal resistors R in parallel: R_effective = R/n = 6/3 = 2 Ω. Alternatively: 1/R = 1/6 + 1/6 + 1/6 = 3/6 → R = 2 Ω. Parallel resistance is always LESS than the smallest individual resistor — here 2 < 6 ✓. This is the most directly tested resistance calculation in AFCAT.

Q2. In household wiring, all electrical appliances are connected in: AFCAT PYQ

(a) Series(b) Parallel(c) Series-parallel combination(d) Neither

✔ Answer: (b) Parallel

All home appliances (fans, bulbs, ACs) are connected in parallel because: (1) each gets the full supply voltage (230V); (2) each can be switched on/off independently; (3) failure of one doesn't affect others. If in series, switching off one would cut off all — and each would get only a fraction of the voltage.

Q3. A 100W and a 60W bulb are connected across the same mains supply. Which one has higher resistance? ⚡ Tricky

(a) 100W bulb(b) 60W bulb(c) Both equal(d) Cannot be determined

✔ Answer: (b) 60W bulb

From P = V²/R → R = V²/P. At the same voltage: R ∝ 1/P. So lower power → higher resistance. R_60W = V²/60 > R_100W = V²/100. The 60W bulb has higher resistance. Students often assume more power means more resistance — the opposite is true when voltage is constant.

Q4. The fuse wire in household wiring is always placed on the: AFCAT PYQ

(a) Neutral wire(b) Earth wire(c) Live wire(d) Any wire

✔ Answer: (c) Live wire

The fuse must be on the live (phase) wire. If placed on the neutral wire, a fault in the appliance could still keep the appliance at high potential even with fuse blown — still dangerous. Placing it on the live wire ensures that when it blows, the appliance is disconnected from the high-potential source. Safety rules require this.

🧠 Quick Memory Chart — PA07 Electricity

⚡ Ohm's Law

V = IR (rearrange: I=V/R; R=V/I)
R ∝ L; R ∝ 1/A
Metals: R ↑ with temp
Semiconductors: R ↓ with temp
Silver: lowest resistivity

📌 Circuits

Series: R adds up; same I
Parallel: R reduces; same V
2 parallel: R₁R₂/(R₁+R₂)
Home: parallel (independent)
Parallel always < smallest R

🔥 Power & Safety

P = VI = I²R = V²/R
At same V: lower R → more P
Joule's: H = I²Rt
Fuse on LIVE wire
India: 230V, 50Hz AC

📝 Practice Exercise

E1. A 6V battery is connected to a 3Ω resistor. Current flowing is:

(a) 18 A(b) 2 A(c) 0.5 A(d) 9 A

E2. Two resistors of 4Ω and 6Ω are in parallel. Effective resistance:

(a) 10 Ω(b) 2.4 Ω(c) 5 Ω(d) 24 Ω

E3. 1 unit of electricity (kWh) equals:

(a) 3600 J(b) 1000 J(c) 3.6 × 10⁶ J(d) 1 J

Answers: E1 → (b) 2 A [I = V/R = 6/3 = 2 A] | E2 → (b) 2.4 Ω [R = 4×6/(4+6) = 24/10 = 2.4 Ω] | E3 → (c) 3.6 × 10⁶ J [1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J]

📝 Mock Tests 🎯 Subject Quiz ✈️ Telegram