Question

Ten identical cells each emf 2V and internal resistance 1Ω are connected in series with two cells wrongly connected. A resistor of 10Ω is connected to the combination. What is the current through the resistor?

• A. 2.4 A
• B. 0.6 A
• C. 1.2 A
• D. 1.8 A

Answer 1

The Correct Option is B

Given:
Total number of cells = 10, 
EMF of each cell = \(2\,V\),
Internal resistance of each cell = \(1\,\Omega\),
External resistance, \( R = 10\,\Omega \),
Number of wrongly connected cells = 2.

Step-by-step Explanation:

Step 1: Determine the effective EMF of the battery combination.

In series, correctly connected cells add their EMFs, whereas wrongly connected cells subtract their EMFs.

Correctly connected cells = \(10 - 2 = 8\)

Net EMF of combination (\(E_{net}\)): \[ E_{net} = (8 \times 2\,V) - (2 \times 2\,V) = 16\,V - 4\,V = 12\,V \]

Step 2: Calculate total internal resistance:

All cells are in series, thus internal resistances add up directly. Total internal resistance (\(r_{total}\)): \[ r_{total} = 10 \times 1\,\Omega = 10\,\Omega \]

Step 3: Find the total resistance of the circuit:

Total resistance (\(R_{total}\)) = Internal resistance + External resistance: \[ R_{total} = r_{total} + R = 10\,\Omega + 10\,\Omega = 20\,\Omega \]

Step 4: Calculate the current (\(I\)) through the resistor using Ohm’s law:

\[ I = \frac{E_{net}}{R_{total}} = \frac{12\,V}{20\,\Omega} = 0.6\,A \]

Final Conclusion:
The current through the resistor is 0.6 A.

Answer 2

1. Calculate the Net EMF (\(E_{\text{net}}\)):
When \(N\) cells are connected in series and \(m\) of them are wrongly connected, the net EMF is given by the EMF of the correctly connected cells minus the EMF of the wrongly connected cells. The number of correctly connected cells is \(N - m = 10 - 2 = 8\). The number of wrongly connected cells is \(m = 2\). \[ E_{\text{net}} = (N - m)E - mE = (N - 2m)E \] Substituting the values: \[ E_{\text{net}} = (10 - 2 \times 2) \times 2\,\text{V} = (10 - 4) \times 2\,\text{V} = 6 \times 2\,\text{V} = 12\,\text{V} \]

2. Calculate the Total Internal Resistance (\(r_{\text{total}}\)):
Internal resistances always add up when connected in series, regardless of the polarity of connection. \[ r_{\text{total}} = N \times r = 10 \times 1\,\Omega = 10\,\Omega \]

3. Calculate the Total Resistance (\(R_{\text{total}}\)):
The total resistance of the circuit is the sum of the total internal resistance and the external resistance. \[ R_{\text{total}} = r_{\text{total}} + R = 10\,\Omega + 10\,\Omega = 20\,\Omega \]

4. Calculate the Current (I):
Using Ohm's law for the entire circuit: \[ I = \frac{E_{\text{net}}}{R_{\text{total}}} \] \[ I = \frac{12\,\text{V}}{20\,\Omega} = \frac{12}{20}\,\text{A} = \frac{3}{5}\,\text{A} = 0.6\,\text{A} \]

Answer: The current through the resistor is 0.6 A. This corresponds to option (B).