How to determine the power of this circuit?

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The discussion revolves around calculating the power of a circuit using nodal analysis, with the textbook providing a power value of 100 W. The user detailed their calculations, arriving at a power output of 98.4 W based on their input current of 4.92 A and a voltage of 20 V. Another participant confirmed the user's calculations were correct, noting their own result of 4.91 A using the same method. Additionally, a suggestion was made to utilize an online circuit simulator for further practice. The conversation highlights the importance of accurate calculations in electrical engineering problems.
Leonid92
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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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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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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