Question

In the circuit shown in Fig. The current through the 10$\Omega$ resistor is:

• A. $\frac{1}{9}A$
• B. $\frac{4}{9}A$
• C. $\frac{2}{3}A$
• D. $\frac{5}{6}A$

Answer 1

Answer: (B)

$\frac{4}{9}A$

Answer 2

The circuit can be analyzed using nodal analysis. Let the potential at the top node connected to the positive terminal of the 12V source be $V_A = 12V$. The ground potential is 0V. Let $V_B$ be the potential at the junction between the two 3$\Omega$ resistors and the 2$\Omega$ resistor. Let $V_C$ be the potential at the junction between the 2$\Omega$ resistor and the 10$\Omega$ resistor.

Applying Kirchhoff's Current Law (KCL) at node B:

The current from node A to node B through the 3$\Omega$ resistor is $I_{AB} = \frac{V_A - V_B}{3} = \frac{12 - V_B}{3}$.

The current from node B to ground through the 3$\Omega$ resistor is $I_{B-ground} = \frac{V_B - 0}{3} = \frac{V_B}{3}$.

The current from node B to node C through the 2$\Omega$ resistor is $I_{BC} = \frac{V_B - V_C}{2}$.

According to KCL at node B, the sum of currents entering the node equals the sum of currents leaving the node:

$I_{AB} = I_{B-ground} + I_{BC}$

$\frac{12 - V_B}{3} = \frac{V_B}{3} + \frac{V_B - V_C}{2}$

Multiplying by 6 to clear the denominators:

$2(12 - V_B) = 2V_B + 3(V_B - V_C)$

$24 - 2V_B = 2V_B + 3V_B - 3V_C$

$24 = 7V_B - 3V_C$ (Equation 1)

Applying KCL at node C:

The current from node B to node C through the 2$\Omega$ resistor is $I_{BC} = \frac{V_B - V_C}{2}$.

The current from node C to ground through the 10$\Omega$ resistor is $I_{C-ground} = \frac{V_C - 0}{10} = \frac{V_C}{10}$.

According to KCL at node C:

$I_{BC} = I_{C-ground}$

$\frac{V_B - V_C}{2} = \frac{V_C}{10}$

Multiplying by 10 to clear the denominators:

$5(V_B - V_C) = V_C$

$5V_B - 5V_C = V_C$

$5V_B = 6V_C$ (Equation 2)

Now we have a system of two linear equations with two variables $V_B$ and $V_C$:

1. $7V_B - 3V_C = 24$
2. $5V_B = 6V_C$

From Equation 2, we can express $V_B$ in terms of $V_C$:

$V_B = \frac{6}{5}V_C$

Substitute this expression for $V_B$ into Equation 1:

$7\left(\frac{6}{5}V_C\right) - 3V_C = 24$

$\frac{42}{5}V_C - 3V_C = 24$

To combine the terms with $V_C$, find a common denominator:

$\frac{42}{5}V_C - \frac{15}{5}V_C = 24$

$\frac{42 - 15}{5}V_C = 24$

$\frac{27}{5}V_C = 24$

$V_C = \frac{24 \times 5}{27} = \frac{8 \times 3 \times 5}{9 \times 3} = \frac{40}{9}V$

The current through the 10$\Omega$ resistor is the current from node C to ground, which is given by Ohm's Law:

$I_{10\Omega} = \frac{V_C - 0}{10} = \frac{V_C}{10}$

$I_{10\Omega} = \frac{40/9}{10} = \frac{40}{9 \times 10} = \frac{40}{90} = \frac{4}{9}A$