Animated Solution for Physics - Current Electricity: In above figure shown, the current in the 10 V battery is close to

Imagine you are an explorer navigating a complex network of rivers. In the world of electronics, these rivers are wires, and the water is the electric current. When circuits have multiple loops, simple Ohm's law isn't enough. We need a heavier tool. Enter Kirchhoff's Voltage Law (KVL), a fundamental principle of energy conservation that will be our compass.

The Loops

To conquer this circuit, we must divide and conquer. Let's split the circuit into two distinct loops. For the left loop, let's assume a current $i_1$ flowing in a clockwise direction. For the right loop, let's assume a current $i_2$ flowing in a counter-clockwise direction.

Why these directions? It's a strategic choice! Because of these directions, both $i_1$ and $i_2$ will flow downwards through the shared $10\Omega$ resistor in the middle. This means the total current there is simply $(i_1+i_2)$, keeping our algebra clean and friendly.

A Clockwise Journey

Let's trace the left loop, starting from the bottom-left corner and marching clockwise. First, we encounter the $20\text{ V}$ battery. We move from the short negative plate to the long positive plate—a step up in potential! So, we write $+20$.

Next, we push through the $5\Omega$ resistor. Since we are moving with the current, we lose energy: $-5i_1$. Down the middle $10\Omega$ resistor, we face the combined current, resulting in a massive drop: $-10(i_1+i_2)$. Finally, we turn left through the $2\Omega$ resistor, dropping by $-2i_1$, and return to our starting point.

Setting the total change to zero gives our first master equation:

Simplifying this beauty, we get:

A Counter-Clockwise Adventure

Now, let's tackle the right loop, moving counter-clockwise from the bottom-right. We climb up the $10\text{ V}$ battery, gaining potential: $+10$. The top wire is a smooth ride with zero resistance. Then, we plunge down the middle $10\Omega$ resistor again, dropping by $-10(i_1+i_2)$.

Finally, we move left through the $4\Omega$ resistor. Since we are moving with our assumed $i_2$, it's a drop: $-4i_2$. Arriving back at the start, we complete the loop:

Combining the terms and dividing by 2, we forge our second master equation:

We now stand before a classic system of two linear equations. Our ultimate prize is $i_2$, the current flowing through the $10\text{ V}$ battery. Let's isolate $i_1$ from the first equation:

Substituting this into our second equation sets the stage for the final calculation:

Multiply the 5 through the numerator, and then multiply the entire equation by 17 to shatter that annoying denominator:

We have our number: $0.21\text{ A}$. But what about that negative sign? In physics, a negative sign is never a mistake; it's a message. It tells us that our initial assumption about the direction was backwards.

We assumed $i_2$ was flowing counter-clockwise, which would mean it travels up through the $10\text{ V}$ battery (from negative to positive). The negative sign reveals the truth: the current actually flows downwards, from the positive terminal to the negative terminal. Thus, the correct answer is $0.21\text{ A}$ from positive to negative.

Thevenin's Shortcut

While KVL is a robust and foolproof method, true masters of circuitry know multiple paths to the summit. Could we have solved this faster? Absolutely. By using Thevenin's Theorem, we could have collapsed the entire left side of the circuit into a single voltage source and a single resistor. This would turn our complex two-loop nightmare into a trivial single-loop calculation. I highly encourage you to try solving it this way—it will make you fall in love with the elegance of circuit analysis!