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.