Find the value of Vx and power dissipated

In summary, the problem involves finding the value of Vx and the power dissipated or provided by the dependent source. The attempt at a solution includes using KVL to get an equation with one unknown, but a second equation is needed to solve for both unknowns. The idea of expressing VA in terms of the 10K resistor and current is suggested as a possible solution.
  • #1
FRZY
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Homework Statement


I'm looking to find the value of Vx and power dissipated or provided by the dependent source.

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Homework Equations



I used KVL to get around the loop, but I'm not sure if that's correct.

The Attempt at a Solution



The main loop I get 10+3Va+10kI = 0.

No clue what to do next. I keep thinking left and right loop, but I already see 4 unknowns 3 equations. Unless I need a junction where I1=-I2. I just don't get it.
Please help, or get my on the right track.
 
Last edited:
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  • #2


Your equation looks okay. One equation, 2 unknowns. You need another one. Consider expressing VA in terms of the 10K resistor and I.
 

Related to Find the value of Vx and power dissipated

1. What is the formula for finding the value of Vx?

The formula for finding the value of Vx is Vx = (Ix * Rx) + (Iy * Ry), where Ix and Iy are the currents flowing through the resistors Rx and Ry, respectively.

2. How do I calculate the power dissipated in a circuit?

The power dissipated in a circuit can be calculated using the formula P = I^2 * R, where I is the current flowing through the circuit and R is the resistance of the circuit.

3. What units are used for Vx and power dissipated?

Vx is measured in volts (V) and power dissipated is measured in watts (W).

4. Can I use Ohm's Law to find the value of Vx and power dissipated?

Yes, Ohm's Law (V = I * R) can be used to find the value of Vx and power dissipated, as long as the values of current and resistance are known.

5. How does the value of Vx affect the power dissipated?

The value of Vx directly affects the power dissipated in a circuit. As Vx increases, the power dissipated also increases. This is because power is directly proportional to voltage (P ∝ V).

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