Consider the circuit shown in the figure: an EMF of 12 V with an in...

Question

Consider the circuit shown in the figure: an EMF of 12 V with an internal resistor of $1\ \Omega$ and an EMF of 6 V with an internal resistor of $0.5\ \Omega$ are connected in parallel between points A and B. What is the potential difference between points A and B?

Accepted Answer

Step 1: Use Millman's theorem for two EMF sources in parallel between the same two nodes.

Step 2: $V_{AB} = \dfrac{\frac{E_1}{R_1}+\frac{E_2}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}} = \dfrac{\frac{12}{1}+\frac{6}{0.5}}{\frac{1}{1}+\frac{1}{0.5}}$.

Step 3: $= \dfrac{12+12}{1+2} = \dfrac{24}{3} = 8\ V$.

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