Animated Solution for Physics - Current Electricity: A $5\text{ V}$ battery with internal resistance $2\Omega$ and a $2\text{ V}$ battery with internal resistance $1\Omega$ are connected to a $10\Omega$ resistor as shown in the figure. The current in the $10\Omega$ resistor is

When faced with a circuit containing multiple batteries and parallel branches, the instinct of many students is to immediately dive into Kirchhoff's Voltage Law (KVL). You might assume two loop currents, say $i_1$ and $i_2$, and write down two simultaneous equations. While this method is perfectly valid, it is often tedious, time-consuming, and highly prone to algebraic errors during an exam.

There is a much more elegant and powerful tool at your disposal: Nodal Analysis.

The Ultimate Shortcut

Nodal Analysis is built entirely on Kirchhoff's Current Law (KCL). The core idea is to find the electric potential at the main junctions (nodes) of the circuit.

In our given circuit, we have three parallel branches connected between two main nodes: $P_1$ at the bottom and $P_2$ at the top. To make our lives incredibly easy, we can arbitrarily choose one of these nodes to be our reference point and assign it a potential of $0\text{ V}$ (grounding it). Let's set $V_{P_1}=0\text{ V}$.

Consequently, let the unknown potential at the top node be $V_{P_2}=V_0$. Our entire mission now boils down to finding this single variable, $V_0$.

According to KCL, the algebraic sum of all currents leaving a junction must be zero. Let's assume that currents $I_1$, $I_2$, and $I_3$ are all flowing out of node $P_2$ and traveling down their respective branches towards $P_1$.

We can express each of these currents using Ohm's Law ($I=\frac{\Delta V}{R}$):

1. The Left Branch: The current $I_1$ travels from $P_2$ to $P_1$. It encounters a $5\text{ V}$ battery. Because it enters the positive terminal and exits the negative terminal, there is a potential drop of $5\text{ V}$. Therefore, $I_1=\frac{V_0-5}{2}$.

2. The Middle Branch: This branch only contains the $10\Omega$ resistor. Therefore, $I_2=\frac{V_0-0}{10}=\frac{V_0}{10}$.

3. The Right Branch (The Trap!): Look very closely at the $2\text{ V}$ battery. Its shorter, negative terminal is facing upwards towards $P_2$. As we travel from $P_2$ to $P_1$, we hit the negative terminal first, meaning we are stepping up in potential across the battery relative to our path. Therefore, the potential difference is $V_0-(-2)$, making $I_3=\frac{V_0+2}{1}$.

Now, we sum them up to form our master KCL equation:

To solve this cleanly, we multiply the entire equation by the least common multiple of the denominators, which is $10$:

Expanding the brackets:

Grouping the terms together yields:

Now that we have the potential at $P_2$, finding the current through the middle $10\Omega$ resistor is trivial:

Converting this to a decimal gives us approximately $0.031\text{ A}$.

Finally, what about the direction? Because our calculated $V_0$ is a positive value ($+\frac{5}{16}\text{ V}$), it confirms that node $P_2$ is at a higher potential than node $P_1$ ($0\text{ V}$). Since conventional current flows from higher to lower potential, the current must flow downwards, from $P_2$ to $P_1$.