How to determine the power of this circuit?

In summary, the conversation discusses a problem in which the power of a circuit needs to be determined using specific equations and resistances. The solution involves using nodal analysis and obtaining a current of 4.92 A and a power of 98.4 W. The answer is slightly different from the one specified in the textbook, but this is due to rounding in the textbook. A helpful resource for practicing similar problems is also provided.
  • #1
Leonid92
45
2
Homework Statement
Determine power of the circuit (Fig. 1), if voltage U = 20 V, resistances of branches: r1 = r4 = 5 Ohm, r2 = r3 = 2 Ohm, r5 = r6 = r7 = 6 Ohm.
Relevant Equations
U(input) = A*U(output) + B*I(output)
I(input) = C*U(output) + D*I(output)
A*D - B*C = 1
In "Homework Equations" field, I wrote general equations for quadripole.
I have doubt about the solution. True answer given in the electrical engineering textbook is P = 100 W.
Here is my solution:
1)
In scheme, I pointed currents and U(output) (Fig. 2).

2)
U = A*U(output) + B*I2
I1 = C*U(output) + D*I2
A = 1 + r1/r3
B = r1 + r2 + (r1*r2)/r3
C = 1/r3
D = 1 + r2/r3

3)
U = A'*U(output) + B'*I6
I5 = C'*U(output) + D'*I6
A' = 1 + r5/r7
B' = r5 + r6 + (r5*r6)/r7
C' = 1/r7
D' = 1 + r6/r7

4)
U(output) = I(output)*r4

5)
U = A*I(output)*r4 + B*I2
I1 = C*I(output)*r4 + D*I2
U = A'*I(output)*r4 + B'*I6
I5 = C'*I(output)*r4 + D'*I6
I(output) = I2 + I6

6)
U = A*(I2 + I6)*r4 + B*I2
I1 = C*(I2 + I6)*r4 + D*I2,
U = A'*(I2 + I6)*r4 + B'*I6
I5 = C'*(I2 + I6)*r4 + D'*I6

7)
A = 3.5
B = 12 Ohm
C = 0.5 S
D = 2
A' = 2
B' = 18 Ohm
C' = (1/6) S
D' = 2

8) Substitute numbers into the system of equations:
20 = 29.5*I2 + 17.5*I6, (1)
I1 = 4.5*I2 + 2.5*I6, (2)
20 = 10*I2 + 28*I6, (3)
I5 = (5/6)*I2 + (17/6)*I6, (4)

(1) - (3):
19.5*I2 - 10.5*I6 = 0
I2 = (21*I6)/39

(3):
20 = 10*(21*I6)/39 + 28*I6
I6 = (130/217) A

I2 = (21*I6)/39 = (10/31) A

(2):
I1 = 4.5*(10/31) + 2.5*(130/217) = 2.95 A

(4):
I5 = (5/6)*(10/31) + (17/6)*(130/217) = 1.97 A

9)
I(input) = I1 + I5 = 4.92 A

10)
Power of the circuit:
P = I(input)*U = 98.4 W.

Is it right solution?

In the textbook, many answers are rounded, so the answer I obtained does not contradict the answer specified in the textbook.
 

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  • #2
Leonid92 said:
Homework Statement: Determine power of the circuit (Fig. 1), if voltage U = 20 V, resistances of branches: r1 = r4 = 5 Ohm, r2 = r3 = 2 Ohm, r5 = r6 = r7 = 6 Ohm.9)
I(input) = I1 + I5 = 4.92 A

10)
Power of the circuit:
P = I(input)*U = 98.4 W.

Is it right solution?

In the textbook, many answers are rounded, so the answer I obtained does not contradict the answer specified in the textbook.
I solved the problem with nodal analysis, and I got I(input)=4.91 A. Your solution looks correct.
 
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  • #3
Hi

I thought I might add this quick and easy online circuit simulator, that would help you when practicing questions:

https://www.falstad.com/circuit/
 
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1. How do I calculate the power of a circuit?

The power of a circuit can be calculated by multiplying the voltage (V) by the current (I). This calculation is represented by the equation P=VI.

2. What units are used to measure power in a circuit?

The unit of power in a circuit is the watt (W). However, milliwatts (mW) and kilowatts (kW) are also commonly used.

3. What is the difference between AC and DC power?

AC (alternating current) power is the type of electricity that is commonly used in households. It can flow in both directions and changes its direction periodically. On the other hand, DC (direct current) power flows in one direction only and is commonly used in electronics and batteries.

4. How does the resistance of a circuit affect its power?

The power of a circuit is inversely proportional to its resistance. This means that as the resistance increases, the power decreases, and vice versa. This relationship is represented by the equation P=V^2/R, where V is the voltage and R is the resistance.

5. Can the power of a circuit be increased?

Yes, the power of a circuit can be increased by increasing the voltage or the current. However, this must be done carefully as increasing either of these parameters beyond the circuit's capacity can cause damage or failure. Additionally, adding more components in parallel can also increase the overall power of a circuit.

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