Question
The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for (i) the current draw from the cell and (ii) the power consumed in the network.
Solution
From the above figure, we can use the horizontal symmetry to deduce the current.
As both resistance in the circuit, as well as the internal resistance in the circuit, have the same resistance 'r'
The resistance of the loop I will be
Because the circuit I and II are same we can say
RII = 2r/3
Combined resistance will be
R = RI + RII
Now this circuit is series with internal resistance 'r'
Resultant resistance will be
The current drawn from the cell will be
I=3E/4r
And power consumed by the network
