Question

Calculate the equivalent resistance in the circuit.

Answer 1

Answer:

56/5 Ω

Explanation:

Equivalent resistance of the right-most 1 Ω and 5 Ω is 6 Ω.

Now that 6 Ω and 4 Ω are in parallel. So,

$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}$

⇒ $R_{eq}$ = 12/5 Ω

Again 12 Ω and 12/5 Ω are in parallel.

$\frac{1}{R_{eq}} = \frac{1}{12} + \frac{5}{12} = \frac{6}{12}$

⇒ 2 Ω

Now the 1 Ω between c and d is in series with the 2 Ω.

∴ $R_{eq} =$ 3 Ω

Now 3 Ω, 6 Ω, 3 Ω are in parallel

$\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{3} = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}$

⇒ $R_{eq}$ = 6/5 Ω

∴ Total equivalent resistance = 10 Ω + 6/5 Ω [As they are in series]

∴ R = 56/5 Ω

Thanks !