Use Kirchhoff’s rules to find the current through $3 \Omega$ resistor in the circuit shown in the figure :
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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.
Current through the 3 ohm resistor by Kirchhoff's rules Step 1: Read the circuit from the figure:
The 3 V cell is in arm AF. Its longer (positive) plate is on the A side.
The 5 V cell is in arm BC. Its longer (positive) plate is on the C side, so its negative plate faces B.
The resistors are 3 \(\Omega\) in BE, 2 \(\Omega\) in CD and 4 \(\Omega\) in FE. Points E and D are joined by a wire.
A and B are joined by a wire, so they are at the same potential. Step 2: Assign currents and apply the junction rule:
Let \(I_1\) leave the positive plate of the 3 V cell, pass through A and reach junction B.
At B it splits into \(I_2\) (through the 5 V cell, C, the 2 \(\Omega\) resistor and D to E) and \(I_3\) (down the 3 \(\Omega\) resistor from B to E).
At E the two currents join again and \(I_1\) returns through the 4 \(\Omega\) resistor to F.
Junction rule at B:
\[
I_1 = I_2 + I_3 \quad \Rightarrow \quad I_2 = I_1 - I_3
\]
Step 3: Loop rule for the left loop A-B-E-F-A:
Go from B to E through 3 \(\Omega\) along \(I_3\): potential drop \(3I_3\).
Go from E to F through 4 \(\Omega\) along \(I_1\): potential drop \(4I_1\).
Go from F to A through the 3 V cell, from its negative plate to its positive plate: rise of 3 V.
\[
3 - 3I_3 - 4I_1 = 0 \quad \Rightarrow \quad 4I_1 + 3I_3 = 3 \qquad \text{(1)}
\]
Step 4: Loop rule for the right loop B-C-D-E-B:
Go from B to C through the 5 V cell, from its negative plate to its positive plate: rise of 5 V.
Go from C to D through 2 \(\Omega\) along \(I_2\): drop \(2I_2\).
Go from E to B through 3 \(\Omega\) against \(I_3\): rise \(3I_3\).
\[
5 - 2I_2 + 3I_3 = 0
\]
Put \(I_2 = I_1 - I_3\):
\[
5 - 2I_1 + 2I_3 + 3I_3 = 0 \quad \Rightarrow \quad 2I_1 - 5I_3 = 5 \qquad \text{(2)}
\]
Step 5: Solve the two equations:
Multiply (2) by 2:
\[
4I_1 - 10I_3 = 10 \qquad \text{(3)}
\]
Subtract (3) from (1):
\[
13 I_3 = 3 - 10 = -7 \quad \Rightarrow \quad I_3 = -\frac{7}{13}\ \text{A} \approx -0.54\ \text{A}
\]
Then \(I_1 = \frac{3 - 3I_3}{4} = \frac{15}{13}\ \text{A}\) and \(I_2 = I_1 - I_3 = \frac{22}{13}\ \text{A}\).
Check in the right loop: \(5 - 2\left(\frac{22}{13}\right) + 3\left(-\frac{7}{13}\right) = \frac{65 - 44 - 21}{13} = 0\). Step 6: Interpret the sign:
A negative \(I_3\) means the real current is opposite to the direction assumed in Step 2.
So the current in the 3 \(\Omega\) resistor flows upward, from E to B. Final Answer:
The current through the 3 \(\Omega\) resistor is \(\frac{7}{13}\ \text{A} \approx 0.54\ \text{A}\), flowing from E to B.
