The output voltage in the following circuit is (Consider ideal diode case):
The output voltage in the following circuit is (Consider ideal diode case):
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- \( -5 \, \text{V} \)
- \( +5 \, \text{V} \)
- \( 10 \, \text{V} \)
- \( 0 \, \text{V} \)
The Correct Option is D
Approach Solution - 1
To determine the output voltage in the given circuit, we need to analyze the circuit considering the ideal diode case. In an ideal diode scenario, the diode conducts perfectly when forward-biased (zero voltage drop across it) and does not conduct at all when reverse-biased (infinite resistance).
- Initially, consider diode \( D_1 \). Since it is forward-biased with \( +5 \, \text{V} \), it will conduct, allowing current to pass through it without any voltage drop.
- Next, examine diode \( D_2 \). The cathode of \( D_2 \) is connected to ground, making the anode potential higher than ground when the circuit is active. This results in \( D_2 \) being reverse-biased.
- As \( D_1 \) is conducting and \( D_2 \) is not conducting, the voltage at the output \( V_{\text{out}} \) is the same as the cathode of the conducting diode \( D_1 \), causing it to be \( +5 \, \text{V} \).
Finally, observe the resistor connected to the output; it is also connected to \( +5 \, \text{V} \). With both connections at the same potential, there is no potential difference, leading the output voltage \( V_{\text{out}} \) to be \( 0 \, \text{V} \).
Thus, the correct answer is:
\( 0 \, \text{V} \)
Approach Solution -2
In the given circuit, we are considering ideal diodes.
The behavior of an ideal diode is:
- It conducts when forward-biased (anode is more positive than cathode).
- It does not conduct when reverse-biased.
Let's analyze the circuit step by step:
1. Diode \( D_1 \) is forward biased because its anode is at \( +5 \, \text{V} \) and its cathode is at \( V_{\text{out}} \).
2. Diode \( D_2 \) is reverse biased because its anode is at ground potential (0V) and its cathode is at \( V_{\text{out}} \). In this configuration:
- \( D_1 \) will conduct, and the output voltage at \( V_{\text{out}} \) will be 0V, since the ideal diode has no voltage drop when it conducts.
- \( D_2 \) will not conduct as it is reverse biased.
Thus, the output voltage \( V_{\text{out}} \) is \( 0 \, \text{V} \).
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