Question

In the given circuit, the current Ix (in mA) is

Hint:

For circuits with dependent sources, systematically apply KCL at critical nodes and use current division rules to find unknown currents.

Answer 1

Correct Answer: 2

To solve for I_x in the given circuit, we will apply Kirchhoff's Current Law (KCL) and the concept of a dependent voltage source.

Step 1: Apply KCL at the top node:

• The incoming current is 5 mA.
• The outgoing currents are through the 1 kΩ resistor (I₀), the dependent source, and the 2 mA current source.
• Thus, KCL gives us: 5 mA = I₀ + I_x + 2 mA.
• Simplifying the equation: I₀ + I_x = 3 mA. (Equation 1)

Step 2: Determine the dependent voltage source:

• The controlled voltage source is given as 1000I₀ volts.
• Ohm’s Law for the 1 kΩ resistor gives V = I₀ × 1 kΩ = 1000 I₀.
• This voltage drop is equal to the drop across the resistor in series with I_x, yielding another equation: (1000 I₀) / 1 kΩ = I_x.
• Simplifying: I_x = 1000 I₀. (Equation 2)

Step 3: Solve the system of equations:

• From Equation 1: I_x = 3 mA - I₀.
• Substitute Equation 2 into this: 1000 I₀ = 3 mA - I₀.
• Simplifying: 1001 I₀ = 3 mA.
• Solving for I₀: I₀ = 3/1001 mA ≈ 0.002997 mA.

Step 4: Calculate I_x:

• Substitute I₀ back into Equation 2: I_x = 1000 × 0.002997 mA.
• Thus, I_x ≈ 2.997 mA.

Verification:

• The computed value of I_x is 2.997 mA.
• Given range is (2, 2 mA), hence I_x fits within this range as expected.

Conclusion: The current I_x is approximately 2.997 mA, within the specified range.