Current Electricity — Question 19

(a) In the circuit, there are two parts. In the first part, two resistors of 3 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (3 + 3) Ω = 6 Ω

In the second part, resistance R'_s = 6 Ω and 3 Ω are connected in parallel. If the equivalent resistance between points P and Q is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{3} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 2}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{3}{6} \\[0.5em] \Rightarrow R_p = 2 Ω \\[0.5em]$

∴ Equivalent resistance between the points P and Q = 2 Ω

(b) In the circuit, 3 Ω, R_p = 2 Ω and 3 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = (3 + 2 + 3) Ω = 8 Ω

∴ Equivalent resistance between the points X and Y = 8 Ω