Question

The current I in the circuit shown is _______.

Answer 1

3.25 x 10^{-3} A

Answer 2

The problem asks us to find the current 'I' in the given circuit. We can solve this using nodal analysis or the superposition theorem. Nodal analysis is generally more straightforward for circuits with multiple sources and interconnected nodes.

1. Define Nodes and Voltages: Let the bottom wire be the ground reference (0 V). Let the node connected to the positive terminal of the 5V voltage source be Node 1. From the diagram, the 5V source is connected between Node 1 and ground, so the potential at Node 1 is $V_1 = 5 \text{ V}$. The current 'I' is the total current flowing out of the 5V source into Node 1. Let the node between the two horizontal 2 kΩ resistors and the vertical 2 kΩ resistor be Node 2, with potential $V_2$. Let the node between the rightmost horizontal 2 kΩ resistor and the current source be Node 3, with potential $V_3$.

2. Apply Kirchhoff's Current Law (KCL) at Node 2 and Node 3:

KCL at Node 2 ($V_2$): The sum of currents leaving Node 2 must be zero. Current from Node 1 to Node 2: $\frac{V_2 - V_1}{2 \text{ k}\Omega}$ Current from Node 2 to Node 3: $\frac{V_2 - V_3}{2 \text{ k}\Omega}$ Current from Node 2 to Ground: $\frac{V_2 - 0}{2 \text{ k}\Omega}$

So, we have: $\frac{V_2 - V_1}{2 \text{ k}\Omega} + \frac{V_2 - V_3}{2 \text{ k}\Omega} + \frac{V_2}{2 \text{ k}\Omega} = 0$ Multiply by $2 \text{ k}\Omega$: $(V_2 - V_1) + (V_2 - V_3) + V_2 = 0$ Substitute $V_1 = 5 \text{ V}$: $(V_2 - 5) + (V_2 - V_3) + V_2 = 0$ $3V_2 - V_3 - 5 = 0$ $3V_2 - V_3 = 5 \quad \text{(Equation 1)}$

KCL at Node 3 ($V_3$): The sum of currents leaving Node 3 must be zero. Current from Node 2 to Node 3: $\frac{V_3 - V_2}{2 \text{ k}\Omega}$ Current from the current source: The arrow indicates current flowing from ground to Node 3, so it's an incoming current of $10^{-3} \text{ A}$. If we consider currents leaving the node as positive, then the current due to the source leaving Node 3 is $-10^{-3} \text{ A}$.

So, we have: $\frac{V_3 - V_2}{2 \text{ k}\Omega} - 10^{-3} = 0$ $\frac{V_3 - V_2}{2000} = 10^{-3}$ $V_3 - V_2 = 2000 \times 10^{-3}$ $V_3 - V_2 = 2 \quad \text{(Equation 2)}$

3. Solve the System of Equations: We have two linear equations with two unknowns ($V_2$ and $V_3$):

1. $3V_2 - V_3 = 5$
2. $-V_2 + V_3 = 2$

Add Equation 1 and Equation 2: $(3V_2 - V_3) + (-V_2 + V_3) = 5 + 2$ $2V_2 = 7$ $V_2 = \frac{7}{2} = 3.5 \text{ V}$

Substitute $V_2 = 3.5 \text{ V}$ into Equation 2: $V_3 - 3.5 = 2$ $V_3 = 2 + 3.5$ $V_3 = 5.5 \text{ V}$

4. Calculate the Current 'I': The current 'I' is the total current leaving the 5V source and entering Node 1. This current splits into two branches:

• Current flowing from Node 1 to Node 2 through the 2 kΩ resistor: $I_{12} = \frac{V_1 - V_2}{2 \text{ k}\Omega}$
• Current flowing from Node 1 to Ground through the 2 kΩ resistor: $I_{1G} = \frac{V_1 - 0}{2 \text{ k}\Omega}$

Therefore, $I = I_{12} + I_{1G}$ $I = \frac{V_1 - V_2}{2000} + \frac{V_1}{2000}$ Substitute $V_1 = 5 \text{ V}$ and $V_2 = 3.5 \text{ V}$: $I = \frac{5 - 3.5}{2000} + \frac{5}{2000}$ $I = \frac{1.5}{2000} + \frac{5}{2000}$ $I = \frac{1.5 + 5}{2000}$ $I = \frac{6.5}{2000}$ $I = \frac{6.5}{2 \times 10^3}$ $I = 3.25 \times 10^{-3} \text{ A}$ Or, $I = 3.25 \text{ mA}$.