Question

A 10V battery of negligible internal resistance is charged by 200V DC supply. If resistance in charging circuit is 38$\Omega$. What is the value of charging current.

Answer 1

5A

Answer 2

The circuit consists of a 200V DC supply, a 38$\Omega$ resistance, and a 10V battery being charged, all connected in series. The 10V battery's emf opposes the charging voltage from the 200V supply.

The net voltage driving the current in the circuit is the difference between the supply voltage and the battery's emf:

$V_{net} = V_{supply} - E_{battery} = 200V - 10V = 190V$.

The total resistance in the circuit is the sum of the charging resistance and the battery's internal resistance. Since the battery's internal resistance is negligible, the total resistance is:

$R_{total} = R_{charging} + r_{battery} = 38\Omega + 0\Omega = 38\Omega$.

Using Ohm's law, the charging current $I$ is given by:

$I = \dfrac{V_{net}}{R_{total}} = \dfrac{190V}{38\Omega} = 5A$.

The value of the charging current is 5A.