Current Electricity — Question 18

In the circuit, there are two parts. In the first 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 second part, the resistance of first part (R'_s = 6 Ω ) and 2 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 3}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{4}{6} \\[0.5em] \Rightarrow R_p = 1.5 Ω \\[0.5em]$

∴ R_p = 1.5 Ω

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

R'_s = (2 + 2) Ω = 4 Ω

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

R''_s = (2 + 2) Ω = 4 Ω

In the third part, the two parts of resistance R'_s = 4 Ω and R''_s = 4 Ω are connected in parallel. If the equivalent resistance between points A and C is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{4} + \dfrac{1}{4} \\[0.5em] \dfrac{1}{R_p} = \dfrac{2}{4} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1}{2} \\[0.5em] \Rightarrow R_p = 2 Ω \\[0.5em]$

∴ Equivalent Resistance between A and C = 2 Ω