In the below circuit, the current in each resistance is
(a) 0.25 A
(b) 0.5 A
(c) 0 A
(d) 1 A

Q: Why did we choose the bottom-left node as $0\text{V}$?

In nodal analysis, you can choose any node as your reference ($0\text{V}$). We chose the bottom-left node simply for convenience. The final physical currents will be the same regardless of where you place the reference.

Q: Why does the potential drop by $2\text{V}$ when moving right?

Look at the battery symbols. The longer line represents the positive terminal, and the shorter line represents the negative terminal. When you move from left to right, you go from positive to negative, which means you are moving to a lower potential.

Q: Can we solve this using loop analysis (KVL)?

Absolutely! You can set up two loop equations. However, because of the symmetry, you will find that the equations simplify to show that the currents are zero. Nodal analysis just gets us to the answer much faster.

Q: Why must the sum of currents leaving the top nodes be zero?

Think of the entire top wire as an isolated island connected to the rest of the circuit only through the three resistors. By conservation of charge, whatever current leaves this island must sum to zero, as charge cannot magically appear or disappear.

Q: What if the resistors had different values, like $1\Omega$, $2\Omega$, and $3\Omega$?

The potential difference across each resistor would still be $V$. The KCL equation would become $\frac{V}{1} + \frac{V}{2} + \frac{V}{3} = 0$. Solving this still gives $V = 0$. So, the current would still be zero regardless of the resistor values!

Question

In the below circuit, the current in each resistance is

Options:

(A) 0.25 A

(B) 0.5 A

(C) 0 A

(D) 1 A

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are looking at a beautifully symmetric circuit. We have three vertical branches, each containing a $1\Omega$ resistor. These branches are connected by top and bottom wires, and interestingly, these connecting wires contain $2	ext{V}$ batteries.

At first glance, this might look like a complex multi-loop circuit that requires solving three simultaneous equations using Kirchhoff's Voltage Law (KVL). But wait! There is a much more elegant way to approach this: **Nodal Analysis**.

### The Master Equation
To use nodal analysis, we need a reference point. Let's arbitrarily choose the bottom-left node and set its potential to $0	ext{V}$. We will call this Node D.

Now, let's walk along the bottom wire from left to right. As we move from Node D to the middle Node E, we cross a $2	ext{V}$ battery from its positive to its negative terminal. This means the potential drops by $2	ext{V}$. So, the potential at Node E is $-2	ext{V}$. Continuing to the right Node F, we cross another identical battery, dropping the potential by another $2	ext{V}$. Thus, Node F is at $-4	ext{V}$.

What about the top wire? We don't know the potential at the top-left node (Node A), so let's call it $V$. Moving right along the top wire, we encounter the exact same arrangement of batteries. Therefore, the potential at the top-middle node (Node B) is $V - 2$, and the potential at the top-right node (Node C) is $V - 4$.

### The Beautiful Symmetry
Now, let's look at the potential difference ($\Delta V$) across each of the three resistors:
- For the left resistor: $\Delta V_1 = V_A - V_D = V - 0 = V$
- For the middle resistor: $\Delta V_2 = V_B - V_E = (V - 2) - (-2) = V$
- For the right resistor: $\Delta V_3 = V_C - V_F = (V - 4) - (-4) = V$

This is the magic of this circuit! Every single resistor experiences the exact same potential difference, $V$.

### Final Calculation
To find the value of $V$, we apply **Kirchhoff's Current Law (KCL)** to the entire top section of the circuit. Since charge cannot accumulate in the top wire, the total current flowing down through the three branches must sum to zero.

$$I_1 + I_2 + I_3 = 0$$

Substituting the currents using Ohm's Law ($I = \frac{\Delta V}{R}$):

$$\frac{V}{1} + \frac{V}{1} + \frac{V}{1} = 0$$

$$3V = 0 \implies V = 0	ext{V}$$

Since the potential difference across each resistor is $0	ext{V}$, the current flowing through each of them is exactly **$0	ext{A}$**. The batteries in the top and bottom wires perfectly balance each other out, creating a state of electrical equilibrium where no current flows through the vertical branches!

Animated Solution for Physics - Current Electricity: In the below circuit, the current in each resistance is

Select Answer:

Visualized Solution

Circuit Analysis

Nodal Analysis Setup

Bottom Node Potentials

Top Node Potentials

Potential Difference Across Resistors

Applying Kirchhoff's Current Law

Final Current

Symmetry and Balance

The Sigma Insight: Kirchhoff's Laws

Analyzing the Setup Imagine you are looking at a beautifully symmetric circuit

The Master Equation To use nodal analysis, we need a reference point

The Beautiful Symmetry

Final Calculation To find the value of $V$ , we apply Kirchhoff's Current Law (KCL) to the entire top section of the circuit

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