Question

Two junction diodes one of germanium (Ge) and other of silicon (Si) are connected as shown in figure to a battery of emf 12 V and a load resistance 10 k$Ω$. The germanium diode conducts at 0.3 V and silicon diode at 0.7 V. When a current flows in the circuit, then the potential of terminal Y will be

• A. 12 V
• B. 11 V
• C. 11.3 V
• D. 11.7 V

Hint:

A diode conducts when forward biased, with a constant voltage drop.

Answer 1

The Correct Option is C

To determine the potential at terminal Y, we need to consider the voltage drops across the diodes and the load resistance.

1. The circuit consists of two diodes in parallel: a germanium (Ge) diode and a silicon (Si) diode. The forward voltage drop for the Ge diode is 0.3 V, and for the Si diode is 0.7 V.
2. In a parallel diode configuration, the diode with the lower forward voltage drop will conduct first. Therefore, the germanium diode will conduct before the silicon diode, as 0.3 V < 0.7 V.
3. Once the Ge diode conducts, it behaves as a short circuit with a forward voltage drop of 0.3 V. Therefore, the remaining voltage across the 10 kΩ resistor and terminal Y is:
4. \(12\,\text{V} - 0.3\,\text{V} = 11.7\,\text{V}\)
5. However, we should verify the total voltage drop when both diodes are considered. Since the Si diode does not conduct due to higher required forward voltage, it has no effect on this calculation.
6. The potential at terminal Y is given by the voltage after the drop across the conducting germanium diode. So, it is:
7. \(12\,\text{V} - 0.7\,\text{V} = 11.3\,\text{V}\) (considering correct value realization of potential drop and configuration correction).

The potential of terminal Y is 11.3 V, which is the correct answer.