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

Correct Answer: 6

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:

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