Power dissipated in circuit: What is the total power dissipated in this circuit?

In summary, the conversation discusses using the node-voltage method to find the total power dissipated in a circuit with given values for current and voltage. The solution involves using nodal analysis and solving for the voltages at each node. The incorrect calculation of V4 caused a discrepancy in the final power dissipated value, but was corrected by fixing V4 to be 0V. The correct power dissipated value is 389 Watts.
  • #1
zealeth
25
0

Homework Statement



Use the node-voltage method to find the total power dissipated in the circuit in the figure if i1 = 2A , i2 = 3A and v1=16V

4-14.jpg


Homework Equations



P = iv
v = iR
P = v^2/R

G = 1/R

[itex]\sum[/itex]G connected to node 1 * v_1 - [itex]\sum[/itex]G between node 1 and 2 * v_2 = Current source into node 1

[itex]\sum[/itex]G connected to node 2 * v_2 - [itex]\sum[/itex]G between node 1 and 2 * v_1 = Current source into node 2

The Attempt at a Solution



Nodal analysis. Going clockwise from the node above the voltage source, each node is labeled 1 through 5.

n93Llvh.png


1:
v_1 = -16

2:
-2 = v_2(1/12+12/25+1/20) - v_3(1/20) - v_4(1/25) + 16(1/12)

3:
5 = -v_2(1/20) + v_3(1/20+1/40) - 0 - v_5(1/40)

4:
0 = 0 - v_2(1/25) + 0 + v_4(1/25+1/40) - v_5(1/40)

5:
-3 = 0 - v_3(1/40) - v_4(1/40) + v_5(1/40+1/40)

Solving, v_2 = -10.8, v_3 = 43.7, v_4 = -18.1, v_5 = -47.2

Power dissipated:
P = [itex]\sum[/itex](v_a - v_b)^2/R
P = (-16+10.8)^2/12 + (-10.8-43.7)^2/20 + (-10.8+18.1)^2/25 + (43.7+47.2)^2/40 + (-18.1+47.2)^2/40
P = 380.6 W (incorrect obviously)

Feedback: "Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures."
 
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  • #2
Hi zealeth! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Check your equation for node 2.

You started by saying let V1=-16, so this means V4 must be 0v, fixed at 0v. You can't later determine V4 to be -18.1v.
 
Last edited by a moderator:
  • #3
NascentOxygen said:
Hi zealeth! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Check your equation for node 2.

You started by saying let V1=-16, so this means V4 must be 0v, fixed at 0v. You can't later determine V4 to be -18.1v.

Node 2 was just a typo I think, but nonetheless I got it (P dissipated = 389 Watts). V_4 being 0 was indeed the problem I was having. Thanks.
 
Last edited by a moderator:

1. What is power dissipation in a circuit?

Power dissipation in a circuit refers to the amount of energy that is converted into heat when an electrical current flows through the circuit. It is a measure of the rate at which energy is used or dissipated in the circuit.

2. How is power dissipation calculated?

Power dissipation can be calculated using the formula P = VI, where P is the power dissipated in watts, V is the voltage across the circuit in volts, and I is the current flowing through the circuit in amperes. Alternatively, it can also be calculated using the formula P = I2R, where R is the resistance in the circuit in ohms.

3. What factors affect power dissipation in a circuit?

The main factors that affect power dissipation in a circuit are the voltage, current, and resistance in the circuit. Higher voltages and currents result in higher power dissipation, while higher resistance results in lower power dissipation. Additionally, the type of material used in the circuit can also affect power dissipation.

4. Why is power dissipation important in circuit design?

Power dissipation is important in circuit design because it can help determine the efficiency and performance of the circuit. Excessive power dissipation can lead to overheating and potential damage to components, while low power dissipation can result in a circuit not functioning properly.

5. How can power dissipation be reduced in a circuit?

Power dissipation in a circuit can be reduced by using components with lower resistance, reducing the voltage or current in the circuit, and improving the overall design and layout of the circuit. Additionally, using materials with higher conductivity and lower heat resistance can also help reduce power dissipation.

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