Question

Consider the circuit shown in the figure. The current I flowing through the 10 $\Omega$ resister is

Answer 1

3/34 A

Answer 2

The problem asks us to find the current 'I' flowing through the 10 Ω resistor in the given circuit. We can solve this using nodal analysis.

1. Define Nodes and Ground: Let the bottom line be the ground (0 V). Let the node between the 1 Ω (top left), 2 Ω (bottom left), and 10 Ω resistors be $V_1$. Let the node between the 10 Ω, 2 Ω (top right), and 1 Ω (bottom right) resistors be $V_2$.

2. Apply Kirchhoff's Current Law (KCL) at Node $V_1$: The current leaving node $V_1$ through the 1 Ω resistor (connected to the 3 V source) is $\frac{V_1 - 3}{1}$. The current leaving node $V_1$ through the 2 Ω resistor (connected to ground) is $\frac{V_1 - 0}{2}$. The current leaving node $V_1$ through the 10 Ω resistor (connected to $V_2$) is $\frac{V_1 - V_2}{10}$.

Sum of currents leaving Node $V_1$ equals zero: $\frac{V_1 - 3}{1} + \frac{V_1}{2} + \frac{V_1 - V_2}{10} = 0$ Multiply the entire equation by 10 to clear the denominators: $10(V_1 - 3) + 5V_1 + (V_1 - V_2) = 0$ $10V_1 - 30 + 5V_1 + V_1 - V_2 = 0$ $16V_1 - V_2 = 30 \quad \text{(Equation 1)}$

3. Apply Kirchhoff's Current Law (KCL) at Node $V_2$: The current leaving node $V_2$ through the 10 Ω resistor (connected to $V_1$) is $\frac{V_2 - V_1}{10}$. The current leaving node $V_2$ through the 1 Ω resistor (connected to ground) is $\frac{V_2 - 0}{1}$. The current leaving node $V_2$ through the 2 Ω resistor (connected to the 3 V source) is $\frac{V_2 - 3}{2}$.

Sum of currents leaving Node $V_2$ equals zero: $\frac{V_2 - V_1}{10} + \frac{V_2}{1} + \frac{V_2 - 3}{2} = 0$ Multiply the entire equation by 10 to clear the denominators: $(V_2 - V_1) + 10V_2 + 5(V_2 - 3) = 0$ $V_2 - V_1 + 10V_2 + 5V_2 - 15 = 0$ $-V_1 + 16V_2 = 15 \quad \text{(Equation 2)}$

4. Solve the System of Linear Equations: We have two equations with two unknowns ($V_1$ and $V_2$):

1. $16V_1 - V_2 = 30$
2. $-V_1 + 16V_2 = 15$

From Equation 1, express $V_2$ in terms of $V_1$: $V_2 = 16V_1 - 30$ Substitute this expression for $V_2$ into Equation 2: $-V_1 + 16(16V_1 - 30) = 15$ $-V_1 + 256V_1 - 480 = 15$ $255V_1 = 495$ $V_1 = \frac{495}{255}$ Simplify the fraction by dividing by 5, then by 3: $V_1 = \frac{99}{51} = \frac{33}{17} \text{ V}$ Now, substitute the value of $V_1$ back into the expression for $V_2$: $V_2 = 16\left(\frac{33}{17}\right) - 30$ $V_2 = \frac{528}{17} - \frac{30 \times 17}{17}$ $V_2 = \frac{528 - 510}{17}$ $V_2 = \frac{18}{17} \text{ V}$

5. Calculate the Current I: The current I flows through the 10 Ω resistor from $V_1$ to $V_2$. $I = \frac{V_1 - V_2}{10}$ $I = \frac{\frac{33}{17} - \frac{18}{17}}{10}$ $I = \frac{\frac{33 - 18}{17}}{10}$ $I = \frac{\frac{15}{17}}{10}$ $I = \frac{15}{17 \times 10}$ $I = \frac{15}{170}$ Simplify the fraction by dividing by 5: $I = \frac{3}{34} \text{ A}$

The current I flowing through the 10 Ω resistor is $\frac{3}{34}$ A.

Explanation of the solution: The circuit is analyzed using nodal analysis. Two nodes, $V_1$ and $V_2$, are identified. KCL equations are formulated for each node, expressing the sum of currents leaving the node as zero. This results in a system of two linear equations. These equations are then solved simultaneously to find the node voltages $V_1$ and $V_2$. Finally, the current $I$ through the 10 Ω resistor is calculated using Ohm's law, $I = (V_1 - V_2) / 10$.

Answer: The current I flowing through the 10 Ω resistor is $\frac{3}{34}$ A.