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(C) In the given circuit,the positive terminal of the $3 	ext{ V}$ battery is connected to the cathode of diode $D_1$ and the anode of diode $D_2$.$1

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$0.03$

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(C) In the given circuit,the positive terminal of the $3 	ext{ V}$ battery is connected to the cathode of diode $D_1$ and the anode of diode $D_2$.$1$. Diode $D_1$: The positive terminal of the battery is connected to the n-side (cathode) of $D_1$. Thus,$D_1$ is reverse-biased and acts as an open circuit (no current flows through this branch).$2$. Diode $D_2$: The positive terminal of the battery is connected to the p-side (anode) of $D_2$. Thus,$D_2$ is forward-biased and acts as a closed switch (ideal diode has zero resistance).$3$. Equivalent Resistance: Since $D_1$ is an open circuit,the current flows only through the branch containing $D_2$ and the $30 \Omega$ resistor,which is in series with the $70 \Omega$ resistor.$R_{eq} = 30 \Omega + 70 \Omega = 100 \Omega$$4$. Current Calculation: Using Ohm's law,$I = \frac{V}{R_{eq}}$$I = \frac{3 	ext{ V}}{100 \Omega} = 0.03 	ext{ A}$

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

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