Q173 / 180 · KCET 2025 · Physics

The circuit shown in figure contains two ideal diodes $D_1$ and $D_2$. If a cell of emf 3 V and negligible internal resistance is connected as shown, then the current through $70 \Omega$ resistance, (in ampere) is Figure

  • A
    0.01
  • B
    0.02
  • Correct answer
    C
    0.03
  • D
    0

Solution

$\begin{aligned} & D_1 \Rightarrow \text { revere biased, } \Rightarrow \text { act as open circuit } \\ & D_2 \Rightarrow \text { Forward biased } \\ & R_{c q}=30+70=100 \Omega\end{aligned}$ $\mathrm{I} \Rightarrow \frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{3}{100}=0.03 \mathrm{~A}$

Hence, the correct answer is option: 'C'.

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