Choose the correct wave form that can represent the voltage across R of the following circuit, assum

Q: Why is the voltage across the resistor zero when the diode is reverse biased?

When the diode is reverse biased, it acts like an open switch. This breaks the circuit, meaning no current can flow ($I = 0$). Since the voltage across a resistor is given by Ohm's Law ($V_R = IR$), zero current means zero voltage!

Q: How do we know the polarity of the 3V battery from the diagram?

In standard circuit diagrams, the longer parallel line represents the positive terminal, and the shorter, thicker line represents the negative terminal. Here, the longer line is on top, meaning the battery opposes the positive half-cycle of the AC source.

Q: Why doesn't option (a) work? It clips the wave at 3V just like we calculated!

Option (a) correctly identifies *when* the diode conducts (only when $V_i > 3\text{V}$). However, it incorrectly assumes $V_R = V_i$ during conduction. Because of the battery, the actual voltage is $V_R = V_i - 3$. Option (a) misses this crucial 3V downward shift, keeping the peak at 10V instead of 7V.

Q: What would the circuit need to look like for option (a) to be correct?

For option (a) to be correct, the 3V battery would need to be removed from the main loop and instead used purely as a reference voltage for a comparator circuit, or the graph would need to plot the voltage across the resistor *and* the battery combined.

Q: Is it common for JEE questions to have no correct options?

It's rare, but it happens! Sometimes questions are dropped or marked as 'bonus' due to errors in the options provided by the examiners. This teaches us a valuable lesson: trust your math and physics principles over the given choices.

Question

Choose the correct wave form that can represent the voltage across $R$ of the following circuit, assuming the diode is ideal one.

Sigma Percentile · Accepted Answer

The problem asks us to find the correct waveform for the voltage across a resistor in a circuit containing an AC source, an ideal diode, and a DC battery. This is a classic clipping circuit problem, but with a slight twist that makes it very interesting!

### Analyzing the Setup

Let's start by looking at the circuit. We have a single loop with four components:
1. An AC voltage source $V_i = 10 \sin \omega t$.
2. An ideal diode $D$.
3. A resistor $R$.
4. A DC battery of $3	ext{V}$.

Notice the polarity of the $3	ext{V}$ battery. The longer line is at the top, meaning its positive terminal is facing the resistor, and it opposes the positive half-cycle of the AC source.

Let's assume a current $I$ flows clockwise through the circuit. Applying Kirchhoff's Voltage Law (KVL) around the loop, we get:
$$V_i - V_D - V_R - 3 = 0$$

Since the diode is ideal, its forward voltage drop $V_D$ is zero when it conducts. Therefore, the voltage across the resistor is:
$$V_R = 10 \sin \omega t - 3$$

### The Condition for Conduction

Here is where we need to be careful. A diode is not just a wire; it's a one-way valve for current. It will only conduct when the current $I$ is positive.

From Ohm's Law, $V_R = IR$. For $I > 0$, we must have $V_R > 0$. 
$$10 \sin \omega t - 3 > 0$$
$$10 \sin \omega t > 3$$

This tells us that the diode only turns on when the AC input voltage exceeds $3	ext{V}$. During this time, the voltage across the resistor is $V_R = 10 \sin \omega t - 3$.

What happens when the AC voltage is less than or equal to $3	ext{V}$? The diode becomes reverse-biased and acts like an open switch. The circuit is broken, the current drops to zero, and consequently, the voltage across the resistor becomes exactly zero.

### Plotting the Output Waveform

Now, let's visualize the output waveform $V_R$ over time. 
- When $V_i \le 3	ext{V}$, the output is flat at $0	ext{V}$.
- When $V_i > 3	ext{V}$, the output follows the sine wave but is **shifted downwards by $3	ext{V}$**.

Because of this downward shift, the peak voltage across the resistor is not $10	ext{V}$. It is:
$$V_{	ext{peak}} = 10 - 3 = 7	ext{V}$$

### The Final Verdict

Let's evaluate the given options based on our findings. 
Option (a) correctly shows the waveform being clipped at the $3	ext{V}$ mark. However, it shows the output voltage perfectly tracking the input voltage above $3	ext{V}$, reaching a peak of $10	ext{V}$. It completely misses the $3	ext{V}$ downward shift caused by the battery!

The other options are even more incorrect, clipping at the wrong voltages or following the wrong half-cycles.

Since none of the graphs show a clipped sine wave with a peak of $7	ext{V}$, **none of the given options are correct**. This is a great reminder to always trust your fundamental physics equations over the provided choices!

Animated Solution for Physics - Semiconductor Electronics: Choose the correct wave form that can represent the voltage across $R$ of the following circuit, assuming the diode is ideal one.

Select Answer:

Visualized Solution

$C i r c u i t A na l y s i s & K V L$

$C o n d i t i o n f or C o n d u c t i o n$

$P l o tt in g t h e O u tp u t W a v e f or m$

$C o m p a r in g w i t h O pt i o n s$

The Sigma Insight: Semiconductor and p-n Junction Diode

Analyzing the Setup

The Condition for Conduction

Plotting the Output Waveform

The Final Verdict

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