The given circuit has two ideal diodes D_1 an | vedclass

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(C) In the given circuit, the $10 \,V$ source is connected to the circuit. Looking at the orientation of the diodes: - Diode $D_1$ has its cathode con

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$2.5$

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(C) In the given circuit, the $10 \,V$ source is connected to the circuit. Looking at the orientation of the diodes: - Diode $D_1$ has its cathode connected to the positive potential side (through $R_1$), making it reverse-biased. Thus, $D_1$ acts as an open circuit (no current flows through $R_2$). - Diode $D_2$ has its anode connected to the positive potential side, making it forward-biased. Thus, $D_2$ acts as a closed switch (short circuit). Therefore, the circuit simplifies to a series combination of the $10 \,V$ battery, resistor $R_1 = 2 \,\Omega$, and resistor $R_3 = 2 \,\Omega$. The total resistance in the circuit is $R_{eq} = R_1 + R_3 = 2 \,\Omega + 2 \,\Omega = 4 \,\Omega$. The current flowing through $R_1$ is given by Ohm's law: $I = \frac{V}{R_{eq}} = \frac{10 \,V}{4 \,\Omega} = 2.5 \,A$.

The given circuit has two ideal diodes $D_1$ and $D_2$ connected as shown in the figure. The current flowing through the resistance $R_1$ will be (in $\,A$)

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