A storage battery of emf 8.0 V and internal resistance 0.5 $\Omega$ is charged by a 120 V dc supply through a series resistor of 15.5 $\Omega$. The charging current in the circuit is

1. While charging, the battery emf opposes the supply, so the net driving voltage is $120 - 8.0 = 112\ \text{V}$. 2. Total circuit resistance is $15.5 + 0.5 = 16.0\ \Omega$. 3. Current $I = 112/16.0 = 7.0\ \text{A}$. 4. Forgetting to subtract the battery emf or to add the internal resistance gives the wrong larger values. _Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.25_