Question

A 5V battery with internal resistance 2Ω and a 2V battery with internal resistance 1Ω are connected to a 10Ω resistor as shown in the figure. The current in the 10Ω resistor is

• A. (A) 0.27A P2 to P1
• B. (B) 0.03A P1 to P2
• C. (C) 0.03A P2 to P1
• D. (D) 0.27A P1 to P2

Answer 1

Answer: (A)

(A) 0.27A P2 to P1

Answer 2

Explanation:
The current flow in the circuit is shown below:Let I will be the current in 10Ω resistor and I1, I2 be the currents in loops ABP2P1A and P2CDP1P2, respectivelyApplying Kirchoff's first law at junction P2,I1 + I2 = IApplying Kirchoff's second law for loop ABP2P1A:10(I1 + I2) + 2I1 - 5 = 0⇒ 12I1 + 10I2 = 5...... (i)Again, applying Kirchoff's second law for loop DCP2P1D:10(I1 + I2) +1I2 - 2 = 0⇒ 10I1 + 11I2 = 2 ...... (ii)Solving equations (i) and (ii), we getI1 = 1.09AI2 = -0.81AThus, the total current in 10Ω resistor isI = I1 + I2 ≃ 0.27A from P2 to P1Hence, the correct option is (A).