Question

The value of current in the 6$\Omega$ resistance is :

• A. 4A
• B. 8A
• C. 10A
• D. 6A

Answer 1

Answer: (C)

10A

Answer 2

The problem can be solved using nodal analysis.

1. Assign Node Potentials: Let the common bottom wire be the reference node with potential $V=0$. Let the junction point between the $20\Omega$, $5\Omega$, and $6\Omega$ resistors be node A, with potential $V_A$.
2. Apply Kirchhoff's Current Law (KCL) at Node A:
   • Current flowing from the $140V$ source through the $20\Omega$ resistor to node A: $I_1 = \frac{140 - V_A}{20}$.
   • Current flowing from node A through the $6\Omega$ resistor to the reference node (0V): $I_2 = \frac{V_A - 0}{6} = \frac{V_A}{6}$.
   • Current flowing from node A through the $5\Omega$ resistor to the positive terminal of the $90V$ source (which is at $90V$ relative to the reference): $I_3 = \frac{V_A - 90}{5}$.
3. Formulate the KCL Equation: Assuming currents $I_1$ enters node A and currents $I_2$ and $I_3$ leave node A, the KCL equation is $I_1 = I_2 + I_3$. $\frac{140 - V_A}{20} = \frac{V_A}{6} + \frac{V_A - 90}{5}$
4. Solve for $V_A$: Multiply the equation by the least common multiple of the denominators (20, 6, 5), which is 60: $60 \left( \frac{140 - V_A}{20} \right) = 60 \left( \frac{V_A}{6} \right) + 60 \left( \frac{V_A - 90}{5} \right)$ $3(140 - V_A) = 10V_A + 12(V_A - 90)$ $420 - 3V_A = 10V_A + 12V_A - 1080$ $420 - 3V_A = 22V_A - 1080$ $420 + 1080 = 22V_A + 3V_A$ $1500 = 25V_A$ $V_A = \frac{1500}{25} = 60 \text{ V}$
5. Calculate the Current through the $6\Omega$ Resistance: The current through the $6\Omega$ resistance is $I_2$: $I_{6\Omega} = \frac{V_A}{6} = \frac{60 \text{ V}}{6 \Omega} = 10 \text{ A}$