Question

With reference to the circuit shown in figure, the potential drop across the 5 $\Omega$ resistor will be.

• A. 1 V
• B. 5 V
• C. 15 V
• D. 10 V

Answer 1

The correct option is B 5 V

The only way to solve this problem is to find the current through the 5$\Omega$ and find IR.
Since there is only one loop, we can find the equivalent EMF and Resistance of the circuit to find the current. Apply the formulae you know.
E=10+5=15V
R=10+5 = 15$\Omega$.
$I=\frac{E}{R}=1A$.
Therefore current through 5$\Omega$ is 1A and potential drop across it is, V=1 $\times$5 = 5V