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Current Electricity

Question

The equivalent resistance across A and B is

2 Ω

3 Ω

4 Ω

5 Ω

Solution

4 Ω

Wheatstone Bridge is an application of Kirchoff's rule.

The equivalent circuit can be redrawn as

We have, $\frac{P}{Q} = \frac{R}{S}$

i.e $\frac{4}{4} = \frac{4}{4}$

So, the given circuit is a balanced Wheatstone's bridge. Hence, the equivalent resistance

R_AB = $\frac{(4 + 4) \times (4 + 4)}{(4 + 4) + (4 + 4)}$

R_AB = $\frac{8 \times 8}{8 + 8}$

= $\frac{64}{16}$

R_AB = 4 Ω