Question

In the given circuit values of \(I_1\), \(I_2\), \(I_3\) are respectively

• A. \(1.364\,\text{A},\ 6.727\,\text{A},\ 5.91\,\text{A}\)
• B. \(1.97\,\text{A},\ 3.56\,\text{A},\ 2.784\,\text{A}\)
• C. \(-0.327\,\text{A},\ 5.28\,\text{A},\ 3.197\,\text{A}\)
• D. \(1.523\,\text{A},\ 4.396\,\text{A},\ 1.63\,\text{A}\)

Hint:

In bridge circuits containing a battery in the middle branch, use the supernode method. First write the battery voltage relation, then apply Kirchhoff's current law to the complete supernode.

Answer 1

The Correct Option is A

Step 1: Assign potentials to the circuit nodes.
Let the right terminal be at zero potential: \[ V_R=0 \] The left terminal is connected to a \(10\,\text{V}\) battery, so \[ V_L=10\,\text{V} \] Let the potentials of the top and bottom junctions of the bridge be \(V_T\) and \(V_B\) respectively.
The central battery is of \(4\,\text{V}\), so \[ V_T-V_B=4 \]

Step 2: Apply Kirchhoff's current law to the supernode.
For the supernode containing \(V_T\) and \(V_B\), \[ \frac{V_T-10}{2}+\frac{V_T-0}{1}+\frac{V_B-10}{1}+\frac{V_B-0}{4}=0 \] \[ \frac{V_T-10}{2}+V_T+V_B-10+\frac{V_B}{4}=0 \] Solving this with \[ V_T-V_B=4 \] we get \[ V_T=\frac{80}{11}\,\text{V} \] and \[ V_B=\frac{36}{11}\,\text{V} \]

Step 3: Calculate \(I_1\).
Current through the \(2\,\Omega\) resistor is \[ I_1=\frac{10-V_T}{2} \] \[ I_1=\frac{10-\frac{80}{11}}{2} \] \[ I_1=\frac{15}{11} \] \[ I_1=1.364\,\text{A} \]

Step 4: Calculate \(I_2\).
Current through the \(1\,\Omega\) resistor in the lower left branch is \[ I_2=\frac{10-V_B}{1} \] \[ I_2=10-\frac{36}{11} \] \[ I_2=\frac{74}{11} \] \[ I_2=6.727\,\text{A} \]

Step 5: Calculate \(I_3\).
Using current balance at the top junction, \[ I_3= \frac{V_T}{1}-I_1 \] \[ I_3=\frac{80}{11}-\frac{15}{11} \] \[ I_3=\frac{65}{11} \] \[ I_3=5.91\,\text{A} \]

Step 6: Final conclusion.
Therefore, \[ \boxed{I_1=1.364\,\text{A},\ I_2=6.727\,\text{A},\ I_3=5.91\,\text{A}} \]