Current Electricity — Question 13

In the circuit, there are four parts. In the first part, three resistors of 1 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (1 + 1 + 1) Ω = 3 Ω

In the second part, three resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R''_s then

R''_s = (2 + 2 + 2) Ω = 6 Ω

In the third part, the two parts of resistance R'_s = 3 Ω and R''_s = 6 Ω and 2 Ω are connected in parallel. If the equivalent resistance is R_p then

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

∴ R_p = 1 Ω

In the fourth part 1 Ω, (R_p = 1 Ω) and 1 Ω are connected in series in between points A and B.

Hence, the equivalent resistance is 1 + 1 + 1 = 3 Ω