Electricity — Solutions
Full worked solutions, with method and final answer for every part.
Question 1
(a)V = IR.
V = 0.40 × 15 = 6.0 V
Question 2
(a)For resistors in parallel: 1/R = 1/R1 + 1/R2, or R = R1R2/(R1+R2).
R = (6.0 × 3.0) ÷ (6.0 + 3.0) = 2.0 Ω
(b)In series, resistances add: total R = 2.0 + 4.0 = 6.0 Ω. Then I = V/R.
Rtotal = 6.0 Ω, I = 12 ÷ 6.0 = 2.0 A
Question 3
(a)Energy (kWh) = power (kW) × time (h).
E = 2 × 3 = 6 kWh
(b)Cost = energy (kWh) × cost per kWh.
Cost = 6 × 8 = ₹48
Question 4
(a)A fuse contains a thin wire that melts and breaks the circuit if the current becomes too large, preventing overheating. Earthing (the earth wire) connects the metal casing of an appliance to the ground, providing a low-resistance path so that if a fault occurs, a large current flows and blows the fuse rather than the casing becoming live.
Fuse: melts to break the circuit if current is too high. Earth wire: provides a safe path to ground, preventing the casing from becoming live
Question 5
(a)I = V/R.
12 ÷ 4 = 3.0 A
Question 6
(a)P = VI.
230 × 2 = 460 W
Question 7
(a)P = I²R.
3² × 10 = 90 W
Question 8
(a)Resistances in series simply add.
5 + 7 + 8 = 20 Ω
Question 9
(a)Energy (kWh) = power (kW) × time (h).
1.5 × 2 = 3 kWh
Question 10
(a)In a series circuit, the same current flows through every component, since there is only one path. In a parallel circuit, current splits between the branches, so the current through each branch can differ, though the total current from the supply equals the sum of the branch currents.
Series: same current throughout (one path). Parallel: current splits between branches
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