JEE Physics 2023 Question with Solution

Given: A circuit with a $10 \, \text{V}$ battery, three $10 \, \Omega$ resistors, and diodes $D_1$, $D_2$, and $D_3$.

Find: The current through the battery.

Step 1: Identify the conducting diodes.

From the given polarity of the battery and the orientation of the diodes, diodes $D_1$ and $D_3$ are forward biased and conduct current, while diode $D_2$ is reverse biased and does not conduct.

Step 2: Reduce the circuit.

With $D_2$ non-conducting, the middle branch is effectively open. The remaining circuit consists of two parallel paths between the battery terminals:

• Top branch: a $10 \, \Omega$ resistor with conducting diode $D_1$
• Bottom branch: a $10 \, \Omega$ resistor with conducting diode $D_3$

These two branches are connected in parallel and are in series with the right-side $10 \, \Omega$ resistor.

Step 3: Find the equivalent resistance.

Step 4: Calculate the current through the battery.

Using Ohm's law,

Since each of the two parallel branches carries equal current,

Step 5: Conclusion.

Therefore, the current through the battery is $1.5 \, \text{A}$. The correct option is A.

Given: The diode orientations decide which branches remain active.

Find: Which branches contribute to the battery current and why.

The key idea is to check forward and reverse bias before combining resistors. A forward-biased diode behaves like a conducting path, while a reverse-biased diode opens the branch. Here, $D_1$ and $D_3$ conduct, so the top and bottom $10 \, \Omega$ branches remain active. Diode $D_2$ blocks current, so the middle $10 \, \Omega$ branch is removed from the circuit.

After that, the problem becomes a resistor network: two equal $10 \, \Omega$ resistors in parallel giving $5 \, \Omega$, followed by one more $10 \, \Omega$ resistor in series. This leads to the total resistance $15 \, \Omega$, and the final current through the battery is $1.5 \, \text{A}$.