Question:
A diode is connected in parallel with a resistance as shown in the figure. The most probable current (I)–voltage (V) characteristic is
A diode is connected in parallel with a resistance as shown in the figure. The most probable current (I)–voltage (V) characteristic is
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Parallel diode-resistor:
Reverse bias ⇒ resistor dominates.
Forward bias ⇒ diode dominates.
Updated On: Nov 5, 2026
- Option a
- Option b
- Option c
- Option d
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
A diode and a resistor \( R \) are connected in parallel across a variable battery of voltage \( V \). We have to choose the most likely graph of the total current \( I \) against \( V \).
Step 2: Key Approach:
In a parallel connection the voltage is the same across both parts and the currents add. So \[ I = \frac{V}{R} + I_{\text{diode}}(V) \]
Step 3: Negative voltage (reverse bias):
The diode blocks current, so \( I_{\text{diode}} \approx 0 \). Only the resistor conducts, and \( I = V/R \). This is a straight line through the origin, in the negative direction.
Step 4: Positive voltage (forward bias):
For small \( V \) the diode still carries almost nothing, so the line continues. After the threshold voltage the diode current grows exponentially. It adds to \( V/R \), so the curve bends upward and rises steeply.
Step 5: Check each graph:
(A) is a straight line for negative \( V \) and a steep curve for positive \( V \). This matches.
(B) levels off at high voltage. Neither a diode nor a resistor saturates like this. So it is wrong.
(C) shows humps and looks like a rectified AC signal. It is not a steady I-V curve. So it is wrong.
(D) shows zero current for negative \( V \). That would be a diode alone. The resistor does conduct in reverse, so it is wrong.
Final Answer:
The graph has a linear part for negative voltage and a steep rise for positive voltage. \[ \boxed{\text{(A)}} \]
A diode and a resistor \( R \) are connected in parallel across a variable battery of voltage \( V \). We have to choose the most likely graph of the total current \( I \) against \( V \).
Step 2: Key Approach:
In a parallel connection the voltage is the same across both parts and the currents add. So \[ I = \frac{V}{R} + I_{\text{diode}}(V) \]
Step 3: Negative voltage (reverse bias):
The diode blocks current, so \( I_{\text{diode}} \approx 0 \). Only the resistor conducts, and \( I = V/R \). This is a straight line through the origin, in the negative direction.
Step 4: Positive voltage (forward bias):
For small \( V \) the diode still carries almost nothing, so the line continues. After the threshold voltage the diode current grows exponentially. It adds to \( V/R \), so the curve bends upward and rises steeply.
Step 5: Check each graph:
(A) is a straight line for negative \( V \) and a steep curve for positive \( V \). This matches.
(B) levels off at high voltage. Neither a diode nor a resistor saturates like this. So it is wrong.
(C) shows humps and looks like a rectified AC signal. It is not a steady I-V curve. So it is wrong.
(D) shows zero current for negative \( V \). That would be a diode alone. The resistor does conduct in reverse, so it is wrong.
Final Answer:
The graph has a linear part for negative voltage and a steep rise for positive voltage. \[ \boxed{\text{(A)}} \]
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