Question

A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 ohms as shown in the figure. Find the value of the current in the circuit.

Answer 1

Since the cells are in opposition,

\(E_{net} = E_1 - E_2\)

= (200 - 10) = 190 V

So, current in the circuit is given by, I = 

\(\frac{E_{net}}{R_{eq}} = \frac{190}{38}\) = 5 A