Question

In the given circuit, calculate the current in each resistance.

Hint:

Always apply KVL loop by loop, and then solve using KCL at junctions.

Answer 1

Step 1: Identify the circuit.
The circuit is a Wheatstone-like bridge with three resistors of $1 \, \Omega$ each and two batteries $E_1 = 2V$, $E_2 = 1V$.
Step 2: Apply Kirchhoff’s Current Law (KCL).
Let currents be $I_1$ through AB, $I_2$ through DC, and $I_3$ through AC. At node A: \[ I_1 = I_2 + I_3 \]
Step 3: Apply Kirchhoff’s Voltage Law (KVL).
Loop (A–B–C–A): \[ - E_1 + I_1(1) + I_3(1) = 0 \quad \Rightarrow \quad -2 + I_1 + I_3 = 0 \] \[ I_1 + I_3 = 2 \quad \quad (1) \] Loop (A–C–D–A): \[ I_3(1) + I_2(1) - E_2 = 0 \quad \Rightarrow \quad I_3 + I_2 - 1 = 0 \] \[ I_3 + I_2 = 1 \quad \quad (2) \]
Step 4: Solve equations.
From (1): $I_1 = 2 - I_3$ From (2): $I_2 = 1 - I_3$ Using KCL: $I_1 = I_2 + I_3$ \[ 2 - I_3 = (1 - I_3) + I_3 \quad \Rightarrow \quad 2 - I_3 = 1 \] \[ I_3 = 1 \] So, $I_1 = 1 \, A$, $I_2 = 0 \, A$, $I_3 = 1 \, A$.
Step 5: Conclusion.
Thus, $I_1 = 1A$, $I_2 = 0A$, $I_3 = 1A$.