MHTCET 17 Apr 2026 Shift 1 Paper

Concept:A reverse-biased diode behaves as an open circuit, while a forward-biased diode contributes its forward resistance to the branch.Explanation:The positive terminal of the battery forward-biases diodes D1 and D2. Diode D3 is reverse-biased, so its infinite backward resistance makes it behave like an open switch. Hence, no current flows through the branch containing D3.For the upper branch, D1 is in series with the $150\ \Omega$ resistor, so its resistance is:$$R_1 = 50 + 150 = 200\ \Omega$$For the middle branch, D2 is in series with the $50\ \Omega$ resistor, so its resistance is:$$R_2 = 50 + 50 = 100\ \Omega$$These two conducting branches are connected in parallel. Their equivalent resistance is:$$R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{200 \times 100}{200 + 100} = \frac{200}{3}\ \Omega$$This parallel combination is in series with the $100\ \Omega$ resistor. Therefore, the total resistance is:$$R_{\text{eq}} = \frac{200}{3} + 100 = \frac{500}{3}\ \Omega$$The current through the $100\ \Omega$ resistor is the total current supplied by the battery:$$I = \frac{V}{R_{\text{eq}}} = \frac{5}{500/3} = \frac{15}{500}\ \text{A} = 0.03\ \text{A} = 30\ \text{mA}$$Answer:The current through the $100\ \Omega$ resistance is $30\ \text{mA}$, which corresponds to option B.