A 2V battery is connected across the points A and B as shown in the figure given below. Assuming that the resistance of each diode is zero in forward bias and infinity in reverse bias, the current supplied by the battery when its positive terminal is connected to A is

(a) 0.2 A
(b) 0.4 A
(c) Zero
(d) 0.1 A

(a) Since diode in upper branch is forward biased and in lower branch is reversed biased. So current through circuit $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}+{\mathrm{r}}_{\mathrm{d}}}$; here ${\mathrm{r}}_{\mathrm{d}}$ = diode resistance in forward biasing = 0 .

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