Question

a battery of internal resistance 4 ohm is connected to the network of resistance as shown. in order to give the maximum power to the network, the value or R should be

Answer 1

The circuit shown in the figure is a balanced Wheatstone bridge. S, as we know in a Wheatstone bridge the wire in the centre of the bridge, will have no current in it so in this case, we can neglect the 6 ohm resistance as no current will flow in that wire.

Now there are two series circuit one with R, R, R and the other has R, R,4R on calculating the equivalent resistance for both we get the values 3R and 6R.

                  ${R_1 = R + R + R = 3R; R_2 = R + R + 4R = 6R}.$

Now $R_1$ and $R_2$ are in parallel.

So, equivalent resistance now becomes 2R.

The circuit can deliver maximum power when external resistance = internal resistance.

So, 2R=4  

                                             R = 2 Ω.