Finding Power using Kirchhoff's rule

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The discussion focuses on calculating the power delivered to resistor R1 in a circuit with given resistances and voltages. Participants suggest using Kirchhoff's rules to find the equivalent resistance of R2 and R3, then applying loop equations to determine the current. After solving the equations, the power delivered to R1 is calculated using the formula P=IV. The final consensus among users is that the power calculated is 1250 W, confirming the theoretical approach aligns with practical results. Overall, the thread emphasizes the importance of systematic problem-solving in circuit analysis.
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Homework Statement



Calculate the power delivered to R1
R1=2.00 Ω, R2=4.00 Ω, R3=4.00 Ω, R4=2.00 Ω, E1=60.0 V and E2=30.0 V.

See current in Attachments:

Homework Equations



P=IV=V^2/R

The Attempt at a Solution



Made R2 and R3 equivalent resistors and then used different loops to solve for current. I end up with a current of 0. HELP!
 

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Can you show your working so we can assist you in finding an answer. Do you know which answers you are supposed to get?
 
I figured it out thanks anyways.

Find the equivalent resistance of R2 and R3. Then use two loops:

I' is the current flowing through the equivalent resistance

1) E1 - R1I1 + ReqI' = 0

2)E2 - R4I4-I'Req = 0

3) I1+I'=I4

sub eqn 3 in 2 and solve for I'

Now sub this in 1 and solve for I1

Now use I1 and R1 to get power
 
Looks good!

What did you get for the Power?
 
I got 1250 W
 
Me too!

Just wanted to check the numbers despite the theory being sound, since sometimes crazy things happen between the theory and the practice.
 
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