Two batteries of EMF 7 V (internal resistance 1 Ω) and 13 V (internal resistance 2 Ω) are connected in parallel to a 12 Ω resistor. The current through the 12 Ω resistor is approximately:
A1.50 A
B2.24 A
C8.5 A
D0.71 A
Answer & Solution
Correct answer: D. 0.71 A
Applying KVL to the two loops (textbook Example 9.2): solving the system $12(I_1+I_2) + I_1 = 7$ and $12(I_1+I_2) + 2I_2 = 13$ gives $I_1 \approx -1.53$ A and $I_2 \approx 2.24$ A. Net current through the 12 Ω: $I = I_1 + I_2 \approx 0.71$ A.
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