Animated Solution for Physics - Current Electricity: In the circuit shown below, the current in the $1\Omega$ resistor is

Nodal Analysis

When faced with a multi-loop circuit like this one, Nodal Analysis is often the most elegant and efficient tool in our arsenal. Instead of writing multiple loop equations using Kirchhoff's Voltage Law (KVL), we can solve the entire circuit by finding the electric potential at a single unknown node.

Let's identify the principal nodes. We have two main junctions where three branches meet: node P at the top and node Q at the bottom. To simplify our math, we can arbitrarily choose one of these nodes as our reference point, or ground. Let's ground node Q, which means we set its potential to exactly $0$ V. Now, our only unknown is the potential at node P, which we will simply call $V$.

With our nodes defined, we apply Kirchhoff's Current Law (KCL) at node P. KCL is a statement of the conservation of charge: the total current entering a junction must equal the total current leaving it.

For consistency, it's a great practice to assume that all currents are leaving the node. If our assumption is wrong for any specific branch, the math will naturally correct us by giving a negative value for that current. So, let's say currents $I_1$, $I_2$, and $I_3$ are leaving node P through the left, right, and middle branches, respectively. Our KCL equation is simply:

Now, we need to express each of these currents in terms of our node potential $V$ using Ohm's Law ($I=\frac{\Delta V}{R}$).

The Left Branch ($I_1$): As we trace the path from P to Q through the left branch, we encounter a $6$ V battery and a $3\Omega$ resistor. Notice the polarity of the battery: we cross from the negative terminal to the positive terminal. This represents a potential gain of $6$ V. Therefore, the effective potential difference driving the current is $V-(-6)-0=V+6$. The current is:

The Right Branch ($I_2$): In the right branch, we have two resistors in series ($2\Omega$ and $3\Omega$), giving a total branch resistance of $5\Omega$. We also cross a $9$ V battery, but this time from the positive terminal to the negative terminal. This is a potential drop of $9$ V. The current is:

The Middle Branch ($I_3$): This is the simplest branch, containing only the $1\Omega$ resistor. The current leaving P is just the potential difference divided by the resistance:

Substituting these expressions back into our KCL equation, we get a single equation with one variable:

Let's group the $V$ terms and move the constants to the right side:

Finding a common denominator for the left side ($15$):

Solving for $V$:

The question asks for the current in the $1\Omega$ resistor, which is exactly our $I_3$.

The magnitude of the current is $0.13$ A. The negative sign is crucial: it tells us that the actual current flows in the opposite direction of our initial assumption. Since we assumed $I_3$ was leaving P (flowing from P to Q), the actual current flows from Q to P.