Solved Sample Paper 5 — Question 15

(a) In the circuit, there are three parts. In the first part two resistors of 10 Ω and 40 Ω are connected in parallel. In parallel, the equivalent resistance is R_p, then

$\dfrac{1}{R_P} = \dfrac{1}{10} + \dfrac{1}{40} \\[0.5em] \dfrac{1}{R_P} = \dfrac{4 + 1}{40} \\[0.5em] \dfrac{1}{R_P} = \dfrac{5}{40} \\[0.5em] \Rightarrow R_P = \dfrac{40}{5} = 8 Ω \\[0.5em]$

In the second part three resistors of 30, 20 and 60 Ω are connected in parallel. In parallel, the equivalent resistance is R'_p, then

$\dfrac{1}{R'_P} = \dfrac{1}{30} + \dfrac{1}{20} + \dfrac{1}{60} \\[0.5em] \dfrac{1}{R'_P} = \dfrac{2 + 3 + 1}{60} \\[0.5em] \dfrac{1}{R'_P} = \dfrac{6}{60} \\[0.5em] \Rightarrow R'_P = \dfrac{60}{6} = 10 Ω \\[0.5em]$

In the third part, R_p and R'_p are in series, the equivalent resistance is R then 8 + 10 = 18 Ω

Hence, total resistance = 18 Ω

(b) From Ohm's Law : V = IR

Substituting the values we get, 12 = I x 18

∴ I = $\dfrac{12}{18}$ = 0.67 A.

Hence, total current = 0.67 A