Question

Two cells are connected in opposition as shown. Cell $E_1$ is of 8 V emf and 2 $\Omega$ internal resistance; the cell $E_2$ is of 2 V emf and 4 $\Omega$ internal resistance. The terminal potential difference of cell $E_2$ is:

Answer 1

Identify the Net Emf in Circuit: Since the cells are connected in opposition, the net emf
$E_{\text{net}} = E_1 - E_2 = 8 - 2 = 6 \, \text{V}$.

Calculate Total Internal Resistance: Total internal resistance $R_{\text{total}} = R_1 + R_2 = 2 + 4 = 6 \, \Omega$.

Determine the Current in Circuit: Using Ohm’s law, the current $I$ in the circuit is:

$I = \frac{E_{\text{net}}}{R_{\text{total}}} = \frac{6}{6} = 1 \, \text{A}$

Calculate Terminal Potential Difference of $E_2$: The potential difference across $E_2$ considering the internal drop is:

$V_{E_2} = E_2 + I \times R_2 = 2 + (1 \times 4) = 6 \, \text{V}$