The circuit shown below contains two ideal diodes, each with a forward resistance of 50 ohms. If the

Q: Why is $D_2$ considered an open circuit?

Because it is reverse-biased. The n-side of $D_2$ is facing the positive terminal of the battery, which means it is at a higher potential than the p-side. This prevents current from flowing through that branch.

Q: What does 'ideal diode with forward resistance' mean?

An absolutely ideal diode has zero resistance when forward-biased and infinite resistance when reverse-biased. Here, the problem specifies it has a forward resistance of $50\,\Omega$, meaning it behaves like a $50\,\Omega$ resistor when conducting, but still has no turn-on voltage drop (like the typical $0.7\text{V}$ for silicon).

Q: Why don't we use the $75\,\Omega$ resistor in the calculation?

Since $D_2$ is in series with the $75\,\Omega$ resistor and it blocks the current, absolutely no current flows through that entire middle branch. Therefore, it doesn't contribute to the equivalent resistance of the active circuit.

Q: How do we know the left side of the battery is positive?

In standard circuit diagrams, a battery or DC voltage source is represented by parallel lines of unequal length. The longer line always represents the positive terminal, and the shorter, thicker line represents the negative terminal.

Q: What if both diodes were forward-biased?

If both were forward-biased (for example, if $D_2$ was flipped to point right), current would flow through both parallel branches. You would first calculate the equivalent resistance of the two parallel branches and then add it to the series $100\,\Omega$ resistor.

Question

The circuit shown below contains two ideal diodes, each with a forward resistance of $50\Omega$. If the battery voltage is 6 V, the current through the $100\Omega$ resistance (in ampere) is

Options:

(A) 0.027

(B) 0.020

(C) 0.030

(D) 0.036

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Let's break down this circuit. We have a $6	ext{V}$ battery powering the system. Notice the long vertical line on the battery symbol? That's our positive terminal on the left. The current wants to flow out from here and split into these two parallel branches containing diodes $D_1$ and $D_2$.

### Diode Biasing: The One-Way Valves
Now, here is the catch. Diodes are like one-way valves. Look at diode $D_1$. Its triangle points to the right, meaning its p-side is connected to the positive terminal. So, $D_1$ is **forward-biased** and will let current pass.

But look at $D_2$. It points to the left! Its n-side faces the positive terminal. This means $D_2$ is **reverse-biased**.

### The Equivalent Circuit
Because $D_2$ is reverse-biased, it acts as an open switch. It completely blocks the current. So, we can just ignore that entire middle branch!

On the other hand, $D_1$ is forward-biased. The problem tells us it's an ideal diode but has a forward resistance of $50\,\Omega$. So, we treat $D_1$ as a $50\,\Omega$ resistor.

### Calculating the Net Resistance
Is this much clear? Now our circuit is just a single, simple loop. The current flows through $D_1$, then the $150\,\Omega$ resistor, and finally the $100\,\Omega$ resistor.

Since they are all in series, we just add their resistances.

$$R_{	ext{net}} = R_{D1} + R_{150} + R_{100}$$

$$R_{	ext{net}} = 50 + 150 + 100 = 300\,\Omega$$

### Final Calculation
Let's substitute the values and get the answer. We know the total voltage is $6	ext{V}$, and the total resistance is $300\,\Omega$. Using Ohm's law, the current $I$ is simply $V$ divided by $R_{	ext{net}}$.

$$I = \frac{V}{R_{	ext{net}}}$$

$$I = \frac{6}{300}$$

Six divided by three hundred simplifies to two divided by one hundred, which is exactly **$0.020	ext{ A}$**. And that is the current flowing through the $100\,\Omega$ resistor. It's a very simple question once you identify the biased states of the diodes.

Animated Solution for Physics - Semiconductors: The circuit shown below contains two ideal diodes, each with a forward resistance of $50Ω$ . If the battery voltage is 6 V, the current through the $100Ω$ resistance (in ampere) is

Select Answer:

Visualized Solution

Circuit Analysis

Diode Biasing

Equivalent Circuit

Net Resistance

Current Calculation

Final Answer

The Way Forward

The Sigma Insight: P-N Junction Diode

Analyzing the Setup Let's break down this circuit

Diode Biasing

The Equivalent Circuit Because $D_{2}$ is reverse-biased, it acts as an open switch

Calculating the Net Resistance Is this much clear? Now our circuit is just a single, simple loop

Final Calculation Let's substitute the values and get the answer

Similar Questions

The circuit shown in the figure contains two diodes each with a forward resistance of $50 Ω$ and with infinite backward resistance. If the battery voltage is $6 V$ , the current through the $100 Ω$ resistance (in ampere) is

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Select Answer:

Visualized Solution

Circuit Analysis

Diode Biasing

Equivalent Circuit

Net Resistance

Current Calculation

Final Answer

The Way Forward

The Sigma Insight: P-N Junction Diode

Analyzing the Setup Let's break down this circuit

Diode Biasing

The Equivalent Circuit Because .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D2​ is reverse-biased, it acts as an open switch

Calculating the Net Resistance Is this much clear? Now our circuit is just a single, simple loop

Final Calculation Let's substitute the values and get the answer

Similar Questions

The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the forward biased diode characteristics shown in the figure, the dynamic resistance at ID​=3 mA will be ......... Ω.

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The Zener diode has a $V_{z} = 30 V$ . The current passing through the diode for the following circuit is ......... mA.