Current Electricity — Question 3

(a) In the circuit, two resistors of 2 Ω each are connected in parallel. In parallel, the equivalent resistance is R_P, then

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

Hence, R_P = 1 Ω

(b) In the circuit, three resistors of 1 Ω, 1 Ω and 2 Ω are connected in parallel. In parallel, the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{R_P} = \dfrac{2 + 2 + 1}{2} \\[0.5em] \Rightarrow R_P = \dfrac{2}{5} Ω \\[0.5em] \Rightarrow R_P = 0.4 Ω$

Hence, R_P = 0.4 Ω

(c) In the circuit two resistors of 1 Ω each are connected in parallel. In parallel, the equivalent resistance is R_P, then

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

Hence, R_P = 0.5 Ω

(d) In the circuit there are three parts. In the first part two resistor of 1 Ω each are connected in series. In series, the equivalent resistance is R'_S, then

R'_S = (1 + 1) Ω = 2 Ω

In the second part two resistor of 1 Ω each are connected in series. In series, the equivalent resistance is R''_S, then

R''_S = (1 + 1) Ω = 2 Ω

In the third part,R'_S and R''_S are in parallel, the equivalent resistance is R_P, then

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

Hence, R_P = 1 Ω

Therefore, on observing the values we can say that (a) and (d) have same equivalent resistances i.e., 1 Ω.