Question

Use Kirchhoff’s rules to find the current through $3 \Omega$ resistor in the circuit shown in the figure :

Hint:

When aggressively applying KVL, always remember that traversing forcefully against the assigned current arrow across a given resistor solidly results in a positive voltage gain ($+IR$), not a drop.

Answer 1

Concept:
• Kirchhoff's Current Law (KCL) firmly states that the total electrical current dynamically entering any junction node strictly equals the total current exiting it.
• Kirchhoff's Voltage Law (KVL) firmly states that the directed algebraic sum of all potential differences (voltage drops and gains) traversing completely around any closed physical loop is precisely zero.

Step 1: Define Currents and established Loops
Let the specific current flowing forcefully out of the 3V battery (driving from F upwards to A) be labeled as $I_1$.
This current $I_1$ securely reaches the top junction node B. Let it physically split into $I_3$ flowing directly downwards securely through the central $3 \Omega$ resistor (from B down to E), and $I_2$ flowing squarely rightwards securely towards node C.
By rigorously applying KCL exactly at junction node B:
\[ I_1 = I_2 + I_3 \implies I_2 = I_1 - I_3 \]

Step 2: Apply KVL securely to the Left Loop
Consider the completely closed physical loop A-B-E-F-A and deliberately traverse it squarely in the clockwise direction.
The potential drops across the middle $3 \Omega$ and bottom $4 \Omega$ resistors, and cleanly rises across the 3V battery (moving precisely from short negative plate to long positive plate):
\[ -3 I_3 - 4 I_1 + 3 = 0 \]
Rearrange to cleanly isolate the constants:
\[ 4 I_1 + 3 I_3 = 3 \quad \text{--- (Equation 1)} \]

Step 3: Apply KVL securely to the Right Loop
Consider the completely closed physical loop B-C-D-E-B and deliberately traverse it squarely in the clockwise direction.
The path goes aggressively through the top 5V battery (from long positive plate to short negative plate, causing a strict drop of 5V), drops across the right $2 \Omega$ resistor, and crucially gains potential securely across the central $3 \Omega$ resistor (since we are traversing forcefully strictly against the assigned direction of $I_3$):
\[ -5 - 2 I_2 + 3 I_3 = 0 \]
Rearrange the terms:
\[ 2 I_2 - 3 I_3 = -5 \]
Substitute the previously defined KCL relation $I_2 = I_1 - I_3$ safely into this equation:
\[ 2(I_1 - I_3) - 3 I_3 = -5 \]
\[ 2 I_1 - 2 I_3 - 3 I_3 = -5 \]
\[ 2 I_1 - 5 I_3 = -5 \quad \text{--- (Equation 2)} \]

Step 4: Solve the resulting System of Linear Equations
We will systematically eliminate $I_1$ to directly solve securely for $I_3$.
Multiply Equation 2 entirely by exactly $2$ to perfectly align the respective coefficients of $I_1$:
\[ 4 I_1 - 10 I_3 = -10 \quad \text{--- (Equation 3)} \]
Now, methodically subtract Equation 3 directly from Equation 1:
\[ (4 I_1 + 3 I_3) - (4 I_1 - 10 I_3) = 3 - (-10) \]
The $I_1$ terms completely cancel out:
\[ 3 I_3 + 10 I_3 = 3 + 10 \]
\[ 13 I_3 = 13 \]
\[ I_3 = 1 \text{ A} \]

Step 5: Conclusion
The defined current variable $I_3$ specifically physically represents the absolute current continuously flowing straight through the central $3 \Omega$ resistor.
Therefore, the exact current flowing directly through the $3 \Omega$ resistor is precisely 1 Ampere, effectively directed downwards from junction B to junction E.