A 5 V battery is connected across the points X and Y. Assume D1 and D2 to be normal silicon diodes.

Q: Why did we completely ignore the $5\ \Omega$ resistor?

Because diode $D_2$ is reverse biased, it acts as an open circuit. No current can flow through an open branch, making the $5\ \Omega$ resistor irrelevant to the current calculation.

Q: What does 'normal silicon diode' mean in the question?

It implies the diode is not ideal. A practical silicon diode requires about $0.7\text{ V}$ to turn on, which appears as a voltage drop across it when forward biased.

Q: What if the question stated the diodes were ideal?

If they were ideal, the forward-biased diode $D_1$ would act as a perfect wire with a $0\text{ V}$ drop. The current would then simply be $I = \frac{5}{10} = 0.5\text{ A}$.

Q: How do we know which side of the diode is the anode?

In the circuit symbol for a diode, the flat triangle points in the direction of conventional current. The wide base of the triangle is the anode (p-side), and the line across the tip is the cathode (n-side).

Q: Could current flow backwards through $D_2$?

No, diodes are essentially one-way valves for current. In reverse bias, the depletion region widens, preventing any significant current flow (except for a negligible leakage current which we ignore in such problems).

Question

A $5	ext{ V}$ battery is connected across the points X and Y. Assume $D_1$ and $D_2$ to be normal silicon diodes. Find the current supplied by the battery, if the positive terminal of the battery is connected to point X.

Options:

(A) $\sim 0.5	ext{ A}$

(B) $\sim 1.5	ext{ A}$

(C) $\sim 0.86	ext{ A}$

(D) $\sim 0.43	ext{ A}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's carefully analyze the given circuit. We are presented with two parallel branches connected between terminals X and Y. The top branch contains diode $D_1$ and a $10\ \Omega$ resistor in series. The bottom branch consists of diode $D_2$ and a $5\ \Omega$ resistor.

The problem states that a $5	ext{ V}$ battery is connected across X and Y, with the positive terminal specifically at X. This means point X is at a higher potential compared to point Y.

Now, let's check the biasing of both diodes. Diode $D_1$ has its p-side (anode) connected to the positive terminal X, so it is **forward biased**. On the other hand, diode $D_2$ has its n-side (cathode) connected to the positive terminal X, making it **reverse biased**.

### The Master Equation

Since these are normal silicon diodes, they are not ideal. A practical forward-biased silicon diode acts like a $0.7	ext{ V}$ battery opposing the current flow. A reverse-biased diode, however, acts as an open circuit.

Because the bottom branch containing $D_2$ is an open circuit, absolutely no current can flow through it. The entire current $I$ from the battery will flow exclusively through the top branch.

Let's apply Kirchhoff's Voltage Law (KVL) in this closed loop. Starting from the battery and moving clockwise, we gain $5	ext{ V}$, then drop $0.7	ext{ V}$ across diode $D_1$, and finally drop $10I$ volts across the resistor. This gives us our master equation:

$$5 - V_{D_1} - I 	imes 10 = 0$$

Substituting the known voltage drop for a silicon diode:

$$5 - 0.7 - 10I = 0$$

### Final Calculation

Now, it's just simple algebra. Subtracting the voltage drop gives us the net voltage available for the resistor:

$$4.3 - 10I = 0$$

Moving the $10I$ to the other side, we get:

$$10I = 4.3$$

Dividing by $10$, we find the total current $I$ supplied by the battery:

$$I = \frac{4.3}{10} = 0.43	ext{ A}$$

This perfectly matches option (d). Always remember to check the polarity of your battery before assuming which branches are active!

Animated Solution for Physics - Semiconductors: A $5 V$ battery is connected across the points X and Y. Assume $D_{1}$ and $D_{2}$ to be normal silicon diodes. Find the current supplied by the battery, if the positive terminal of the battery is connected to point X.

Select Answer:

Visualized Solution

Circuit Analysis

Battery Connection

Biasing Check

Equivalent Models

Current Flow

Applying KVL

Solving for $I$

Final Current

Food for thought

The Sigma Insight: P-N Junction Diode

Analyzing the Setup

The Master Equation

Final Calculation

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