Current Electricity — Question 17

In the circuit, there are three parts. In the first part, two resistors of 3 Ω, 2 Ω are connected in series. If the equivalent resistance is R'_s then

R'_s = 3 + 2 = 5 Ω

In the second part, two resistors of 6 Ω, 4 Ω are connected in series. If the equivalent resistance is R''_s then

R''_s = 6 + 4 = 10 Ω

In the third part R'_s, 30 Ω and R''_s are connected in parallel. If the equivalent resistance is R_p then

$\dfrac{1}{R_P} = \dfrac{1}{R'_s} + \dfrac{1}{30} + \dfrac{1}{R''_s} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{5} + \dfrac{1}{30} + \dfrac{1}{10} \\[0.5em] \dfrac{1}{R_P} = \dfrac{6 + 1 + 3}{30} \\[0.5em] \dfrac{1}{R_P} = \dfrac{10}{30} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{3} \\[0.5em] \Rightarrow R_P = 3 Ω$

Hence, R_P = 3 Ω