Question

Find the power delivered by the sources in the circuit shown in figure :

• A. 200 W
• B. 250 W
• C. 500 W
• D. 300 W

Answer 1

The correct option is D 300 W

Equivalent emf of the batteries $E_{eq}=E_1-E_2=50-20=30$ V

Equivalent resistance of the circuit $R_{eq}=R_1+R_2=1+2=3\Omega$

Thus power delivered $P=\frac{E_{eq}^2}{R_{eq}}=\frac{{30}^2}{3}=300W$