Q174 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
A closed loop consists of a $15\ \text{V}$ battery, a $5\ \text{V}$ battery (opposing the first), and a $25\ \Omega$ resistor. The current through the loop is
- A$0.8\ \text{A}$
- B$0.2\ \text{A}$
- C$1.2\ \text{A}$
- Correct answerD$0.4\ \text{A}$
Solution
The net electromotive force in the loop is the difference between the two opposing batteries.
$E_{\text{net}} = 15 - 5 = 10\ \text{V}$
The total resistance in the loop is $R = 25\ \Omega$.
Using Ohm's law, the current $I$ is given by:
$I = \dfrac{E_{\text{net}}}{R} = \dfrac{10}{25} = 0.4\ \text{A}$
Answer: $0.4\ \text{A}$
$E_{\text{net}} = 15 - 5 = 10\ \text{V}$
The total resistance in the loop is $R = 25\ \Omega$.
Using Ohm's law, the current $I$ is given by:
$I = \dfrac{E_{\text{net}}}{R} = \dfrac{10}{25} = 0.4\ \text{A}$
Answer: $0.4\ \text{A}$
Hence, the correct answer is option: 'D'.