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

The problem asks for the terminal potential difference across the cell with emf \(E_2\). The circuit consists of two cells connected in opposition in a simple loop.

Concept Used:

1. Net EMF in a Loop: When two cells are connected in series, the net EMF is the sum of their individual EMFs if they are connected in the same direction, and the difference if they are connected in opposition. In this case, the cells are in opposition, so the net EMF will drive the current in the direction of the larger EMF.

2. Ohm's Law for a Complete Circuit: The current (\(I\)) flowing in a simple loop is given by the ratio of the net EMF to the total resistance in the circuit.

\[ I = \frac{E_{net}}{R_{total}} \]

3. Terminal Potential Difference of a Cell: The terminal potential difference (\(V_T\)) across a cell is the potential difference between its positive and negative terminals. - If the cell is discharging (supplying current), \(V_T = E - Ir\). - If the cell is charging (current is being forced into its positive terminal), \(V_T = E + Ir\).

Step-by-Step Solution:

Step 1: Identify the given values and the configuration of the circuit.

• EMF of the first cell, \(E_1 = 8 \, \text{V}\).
• Internal resistance of the first cell, \(r_1 = 2 \, \Omega\).
• EMF of the second cell, \(E_2 = 2 \, \text{V}\).
• Internal resistance of the second cell, \(r_2 = 4 \, \Omega\).

The cells are connected in opposition. Since \(E_1 > E_2\), the first cell will dominate and drive the current in the circuit. The direction of the current will be counter-clockwise, from the positive terminal of \(E_1\) to its negative terminal.

Step 2: Calculate the net EMF and total resistance of the circuit.

The net EMF, since the cells are in opposition, is:

\[ E_{net} = E_1 - E_2 = 8 \, \text{V} - 2 \, \text{V} = 6 \, \text{V} \]

The total resistance in the circuit is the sum of the internal resistances, as they are in series:

\[ R_{total} = r_1 + r_2 = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega \]

Step 3: Calculate the current flowing in the circuit.

Using Ohm's law for the complete circuit:

\[ I = \frac{E_{net}}{R_{total}} = \frac{6 \, \text{V}}{6 \, \Omega} = 1 \, \text{A} \]

Step 4: Determine the terminal potential difference across cell \(E_2\).

The current of 1 A flows in a counter-clockwise direction. This means the current flows from point C to point B, entering the positive terminal of cell \(E_2\) and leaving through its negative terminal. This indicates that cell \(E_2\) is being charged by cell \(E_1\).

For a cell being charged, the terminal potential difference (\(V_{T2}\)) is given by:

\[ V_{T2} = E_2 + I r_2 \]

Final Computation & Result:

Substitute the values of \(E_2\), \(I\), and \(r_2\) into the formula:

\[ V_{T2} = 2 \, \text{V} + (1 \, \text{A})(4 \, \Omega) \] \[ V_{T2} = 2 \, \text{V} + 4 \, \text{V} = 6 \, \text{V} \]

The terminal potential difference of cell \(E_2\) is 6 V.

Answer 2

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:

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