Question

Solve question 11 please.

Answer 1

It is simple.

The given circuit is a balanced Wheatstone bridge. So the wire containing 6 ohm will be eliminated.

R,R,R are in series while R,R and 4R are in series.

So, R' = R + R + R = 3R
R'' = R + R + 4R = 6R

R' and R'' are in parallel

Effective Resistance = 1/3R + 1/6R = 3/6R = 6R/3 = 2R

For delivering maximum power, internal resistance = external resistance
Given, internal resistance = 4 ohm

So, 2R = 4
R = 2 ohm --> your required answer

Hope This Helps You!