When the switch S in the circuit shown is closed, then the value of current i will be
(a) 4 A
(b) 3

Q: Why did we assume currents $i_1$ and $i_2$ are flowing towards the junction? What if they are actually flowing away?

The initial direction you assume for the currents is completely arbitrary! As long as you write your KCL equation consistently based on your assumption, the math will take care of the rest. If a current is actually flowing in the opposite direction to what you assumed, your final calculated value for that current will simply come out negative. The magnitude will still be perfectly correct.

Q: Can I solve this problem using Kirchhoff's Voltage Law (KVL) loop equations instead?

Absolutely. You could define two loops (e.g., the left loop and the right loop) and write two simultaneous KVL equations. However, solving two simultaneous equations is generally more time-consuming and prone to algebraic errors than solving the single linear equation we get from Nodal Analysis. For circuits with multiple parallel branches, Nodal Analysis is almost always the faster and smarter choice.

Q: What exactly does the ground symbol mean at the bottom of the circuit?

The ground symbol represents a reference point in the circuit where the electrical potential is defined to be exactly $0\text{ V}$. It's like choosing sea level as the zero point for measuring altitude. By setting this point to $0\text{ V}$, it makes calculating the potential differences across the other components much simpler.

Q: Why did we multiply the entire equation by 4 in the middle of the calculation?

We multiplied by 4 because it is the Least Common Multiple (LCM) of the denominators (2 and 4) in our equation. Doing this clears all the fractions in a single step, turning it into a simple linear equation with integers. It's a standard algebraic trick to make the calculation faster and reduce the chance of making a silly mistake with fractions.

Q: What would happen if the switch S was kept open?

If the switch $S$ is open, the central branch is broken, meaning it has infinite resistance. Therefore, no current can flow through it, so $i$ would be exactly $0\text{ A}$. The circuit would then just be a single large loop with the $20\text{ V}$ and $10\text{ V}$ sources opposing each other.

Question

When the switch $S$ in the circuit shown is closed, then the value of current $i$ will be

Options:

(A) 4 A

(B) 3 A

(C) 2 A

(D) 5 A

Sigma Percentile · Accepted Answer

### Mastering Nodal Analysis: Solving Multi-Source Circuits

Imagine you are an electrical engineer tasked with finding the exact current flowing through a specific wire in a complex network. When a circuit has multiple voltage sources, traditional series and parallel reduction techniques often fall short. This is where **Kirchhoff's Current Law (KCL)**, specifically applied through **Nodal Analysis**, becomes your ultimate superpower.

Let's break down this classic JEE Main problem step-by-step and see how elegantly Nodal Analysis slices through the complexity.

### Analyzing the Setup

Look closely at the given circuit. We have two distinct voltage sources: a $20	ext{ V}$ source on the left and a $10	ext{ V}$ source on the right. In the middle, there is a switch $S$. Initially, the switch is open, meaning no current flows through the central branch.

As soon as we close switch $S$, the circuit is completed. Our objective is to find the value of the current $i$ flowing downwards through the central $2\ \Omega$ resistor to the ground.

### The Master Equation: Setting up KCL

To apply Nodal Analysis, we first need to identify our principal node—the junction where multiple branches meet. Let's call the top central junction point $C$ and assume its electrical potential is $V_C$.

We know the potentials at the other ends of the branches:
*   The left branch is connected to the positive terminal of the $20	ext{ V}$ battery, so $V_A = 20	ext{ V}$.
*   The right branch is connected to the positive terminal of the $10	ext{ V}$ battery, so $V_B = 10	ext{ V}$.
*   The bottom branch is connected to the ground, which by convention is at $0	ext{ V}$.

Now, let's assume currents $i_1$ and $i_2$ are flowing *towards* junction $C$ from the left and right sources, respectively. The current $i$ is flowing *away* from junction $C$ towards the ground.

According to **Kirchhoff's Current Law (KCL)**, the total current entering a junction must equal the total current leaving it. Therefore, we can write our master equation:

$$i_1 + i_2 = i$$

### Translating Currents to Potentials

Using Ohm's Law ($i = \frac{\Delta V}{R}$), we can express each of these currents in terms of the node potentials and the branch resistances.

*   For the left branch: $i_1 = \frac{20 - V_C}{2}$
*   For the right branch: $i_2 = \frac{10 - V_C}{4}$
*   For the central branch: $i = \frac{V_C - 0}{2}$

Substituting these into our KCL equation, we get the raw setup:

$$\frac{20 - V_C}{2} + \frac{10 - V_C}{4} = \frac{V_C}{2}$$

### The Final Calculation

I know this equation might look a bit messy with the fractions, but let's take a breath and simplify it. To clear the denominators, we can multiply the entire equation by the least common multiple, which is $4$.

$$2(20 - V_C) + (10 - V_C) = 2V_C$$

Now, let's expand the brackets carefully. Watch out for the minus signs!

$$40 - 2V_C + 10 - V_C = 2V_C$$

Rearranging the terms to group all the $V_C$ variables on one side and the constants on the other:

$$40 + 10 = 2V_C + 2V_C + V_C$$

$$50 = 5V_C$$

Solving for the node potential, we find:

$$V_C = 10	ext{ V}$$

We have successfully found the potential at the central junction! But we are not done yet. The question asks for the current $i$. We simply plug $V_C$ back into our expression for $i$:

$$i = \frac{V_C}{2} = \frac{10}{2} = 5	ext{ A}$$

And there we have it! The current flowing through the switch is **$5	ext{ A}$**. Nodal analysis transformed a potentially confusing multi-loop problem into a straightforward single-variable linear equation. Master this technique, and you'll be solving complex circuits in seconds.

Animated Solution for Physics - Current Electricity: When the switch $S$ in the circuit shown is closed, then the value of current $i$ will be

Select Answer:

Visualized Solution

Initial Circuit Setup

Closing the Switch

Assigning Node Potentials

Assuming Current Directions

Applying KCL

Substituting Ohm's Law

Simplifying the Equation

Expanding Brackets

Rearranging Terms

Solving for Node Potential

Calculating Final Current

The Way Forward

The Sigma Insight: Kirchhoff's Laws

Mastering Nodal Analysis

Analyzing the Setup

The Master Equation

Translating Currents to Potentials

The Final Calculation

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Animated Solution for Physics - Current Electricity: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } When the switch S in the circuit shown is closed, then the value of current i will be