Two identical p-n junctions may be connected in series with a battery in three ways. The potential d

Q: Why does the potential drop depend on the biasing?

A p-n junction is a non-linear device. Its resistance changes drastically depending on whether it is forward-biased (low resistance) or reverse-biased (high resistance). Thus, the potential drop across it will change accordingly.

Q: In circuit-3, since both are reverse biased, won't the current be zero?

The current will be very small (leakage current), but not exactly zero. Since both diodes are identical and in series, this same small current flows through both, and they offer the same high reverse resistance, leading to equal potential drops.

Q: What if the diodes were not identical?

If the diodes were not identical, even with the same biasing, their resistances (forward or reverse) would be different. In that case, the potential drops would not be equal.

Q: How do we quickly identify forward and reverse biasing?

Look at the battery terminals. If the p-side (anode) is connected to the higher potential (positive terminal) relative to the n-side (cathode), it is forward-biased. If the n-side is at a higher potential, it is reverse-biased.

Q: Can we apply Kirchhoff's Voltage Law (KVL) here?

Absolutely! KVL states that the sum of potential drops across the diodes equals the battery voltage. Since $V = I R$, and $I$ is the same for both (series connection), $V_1 = V_2$ only if $R_1 = R_2$.

Question

Two identical $p$-$n$ junctions may be connected in series with a battery in three ways. The potential drops across the two $p$-$n$ junctions are equal in

Options:

(A) circuit-1 and circuit-2

(B) circuit-2 and circuit-3

(C) circuit-3 and circuit-1

(D) circuit-1 only

Sigma Percentile · Accepted Answer

### Understanding the Core Concept

Imagine water flowing through two pipes connected one after the other. If the pipes are identical, the pressure drop across each pipe will be exactly the same. Similarly, in an electrical circuit, when two identical components are connected in series, the same current flows through both of them. According to Ohm's law ($V = I R$), the potential drop across each component will be equal only if their resistances are equal.

A $p$-$n$ junction diode is a non-linear device. Its resistance is highly dependent on how it is connected to the battery, a concept known as **biasing**.
- When a diode is **forward-biased** (the $p$-side is connected to the positive terminal and the $n$-side to the negative terminal), it offers a very low resistance.
- When a diode is **reverse-biased** (the $n$-side is connected to the positive terminal and the $p$-side to the negative terminal), it offers a very high resistance.

Therefore, for two identical diodes in series to have the same resistance, they must be biased in the exact same way. Let's analyze the three given circuits based on this principle.

### Analyzing the Circuits

**Circuit-1:**
In the first circuit, the left diode has its $p$-side facing the positive terminal of the battery, making it forward-biased. However, the right diode has its $n$-side facing the positive terminal, making it reverse-biased. Since one diode offers low forward resistance and the other offers high reverse resistance ($R_1 
eq R_2$), the potential drops across them will be unequal ($V_1 
eq V_2$).

**Circuit-2:**
In the second circuit, both diodes have their $p$-sides facing the positive terminal of the battery. This means both diodes are forward-biased. Since they are identical and biased in the same way, they offer the exact same forward resistance ($R_1 = R_2$). Consequently, the potential drops across them will be perfectly equal ($V_1 = V_2$).

**Circuit-3:**
In the third circuit, both diodes have their $n$-sides facing the positive terminal of the battery. This means both diodes are reverse-biased. Even though they offer a very high resistance, the resistance of both identical diodes will be the same ($R_1 = R_2$). Thus, the potential drops across them will also be equal ($V_1 = V_2$).

### The Final Verdict

From our analysis, it is clear that the identical diodes offer the same resistance in Circuit-2 and Circuit-3 because they share the same biasing state. Therefore, the potential drops across the two $p$-$n$ junctions are equal in **circuit-2 and circuit-3**.

Animated Solution for Physics - Semiconductors: Two identical $p$ - $n$ junctions may be connected in series with a battery in three ways. The potential drops across the two $p$ - $n$ junctions are equal in

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Visualized Solution

$V_{1} = V_{2}$

$Condition for R_{1} = R_{2}$

$Circuit-1 Analysis$

$Circuit-2 Analysis$

$Circuit-3 Analysis$

$Conclusion$

The Sigma Insight: P-N Junction Diode

Understanding the Core Concept

Analyzing the Circuits

The Final Verdict

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V1​=V2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for R1​=R2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit-1 Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit-2 Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit-3 Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion