Animated Solution for Physics - Current Electricity: In the given circuit, the cells have zero internal resistance. The currents (in Ampere) passing through resistances $R_1$ and $R_2$ respectively are

When faced with a multi-loop circuit containing several batteries, the immediate instinct of many students is to dive straight into Kirchhoff's Voltage Law (KVL). While KVL is a powerful tool, writing out multiple loop equations, assigning current variables, and solving simultaneous linear equations can be incredibly time-consuming and prone to silly algebraic mistakes.

There is a much more elegant and faster way to approach such problems: The Node Voltage Method.

The first step in node voltage analysis is to identify a continuous wire that connects to multiple branches and declare it as our reference ground. In this circuit, the entire bottom wire connects the left, middle, and right branches together. By setting this bottom wire to $0\text{ V}$, we create a solid foundation to determine the electric potential at every other point in the circuit.

Once the ground is established, we can read the voltages at the top nodes directly from the batteries:

1. Node B (Top Middle): The middle branch contains a $10\text{ V}$ battery with its negative terminal connected to our $0\text{ V}$ ground. Therefore, the potential at Node B is exactly $10\text{ V}$. 2. Node C (Top Right): Similarly, the right branch has an identical $10\text{ V}$ battery connected to ground. This forces Node C to also be at $10\text{ V}$. 3. Node A (Top Left): The left branch is simply a straight wire connecting Node A directly to the ground. Thus, Node A is at $0\text{ V}$.

With the node voltages clearly defined, finding the currents is just a matter of applying Ohm's Law ($I=\frac{\Delta V}{R}$) to each resistor.

For the first resistor $R_1$, it is connected between Node B ($10\text{ V}$) and Node A ($0\text{ V}$):

Now, let's look at the second resistor $R_2$. It is connected between Node B and Node C. However, we already established that both of these nodes are sitting at exactly $10\text{ V}$!

Because there is absolutely no potential difference across $R_2$, there is no electrical "pressure" to push charges through it. The resistor is effectively short-circuited by the parallel batteries.

By taking a moment to analyze the node potentials before writing any equations, we bypassed all the heavy algebra and arrived at the answer ($0.5\text{ A}$ and $0\text{ A}$) in mere seconds. Work smart, not hard!