Finding Equivalent Resistance, Current, and Voltage across a resistor

AI Thread Summary
The discussion focuses on calculating equivalent resistance, current, and voltage in a circuit with four resistors. The correct equivalent resistance was confirmed, but the initial answers for current (i1) and voltage across the 60-ohm resistor were deemed incorrect. Using Kirchhoff's Voltage Law, the correct current i1 was recalculated to be 3 A, while the voltage across the 60-ohm resistor was confirmed to be 120 V. A suggestion was made to redraw the circuit for clarity, particularly when dealing with diagonally placed resistors. Overall, the conversation emphasizes the importance of accurate application of circuit laws for solving electrical problems.
rugerts
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Homework Statement
A) What's equivalent resistance of all four resistors?
B) What is current value of i1?
C) What is voltage across 60 ohm resistor?
Relevant Equations
Parallel connections add like (1/Req) = (1/R1+1/R2)
Series add normally R1+R2+ ...
Current divider: I = Isource*(Req/Ri)
Voltage divider: V = Vsource*(Ri/Req)
1569385726645.png

1569385736859.png
 
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rugerts said:
Homework Statement: A) What's equivalent resistance of all four resistors?
B) What is current value of i1?
C) What is voltage across 60 ohm resistor?
Homework Equations: Parallel connections add like (1/Req) = (1/R1+1/R2)
Series add normally R1+R2+ ...
Current divider: I = Isource*(Req/Ri)
Voltage divider: V = Vsource*(Ri/Req)

View attachment 250165
View attachment 250166
Answer c is wrong.
 
Hi Rugerts.

Answer A is correct. But answers for B and C are both incorrect.
(Answer for B is asking for current I1)

I1 + I2 = Is
where I1 and I2 are both unknown, and Is is the 5 amps you correctly calculated.

The method I learned was then to write an equation for the voltage rise and drops as you "walk" around a closed loop from the battery's positive terminal to its negative terminal. Are you familiar with that method?
 
CPW said:
Hi Rugerts.

Answer A is correct. But answers for B and C are both incorrect.
(Answer for B is asking for current I1)

I1 + I2 = Is
where I1 and I2 are both unknown, and Is is the 5 amps you correctly calculated.

The method I learned was then to write an equation for the voltage rise and drops as you "walk" around a closed loop from the battery's positive terminal to its negative terminal. Are you familiar with that method?
Are you saying to use Kirchhoff's Voltage Law to find i1? In doing so, I get: -160+ 40i1 + 8i1 = 0 -> i1 = 3.33 A. Could I use a voltage divider to find the voltage across the 60 ohm resistor or no? FYI the loop was the outermost one.
Doing this for the 60 ohm resistor and choosing a loop as the lower left hand triangle, I get a voltage across 60 ohm resistor to be 120 V.
 
Hi again.

Ok. The loop rule I mentioned is what you have obviously been taught (Kirchhoff's Voltage law). I see your equation:
-160 + 40i1 + 8i1 = 0.

But you made an easy mistake in that equation.
The current throught the 8 ohm resistor is not i1, but rather it is iS = 5 A (the same as the current through the battery). Make sense?

With that equation fixed, you should be able to correctly calculate i1 to be 3 A.

I agree with your calculation of the voltage across the 60 ohm resistor: 120V.

One thing that helps me with circuit problems when a resistor is shown diagonally, is to re-draw it without the diagonal.
 

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