Finding Equivalent Resistance, Current, and Voltage across a resistor

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SUMMARY

The discussion focuses on calculating equivalent resistance, current (i1), and voltage across a 60-ohm resistor in a circuit involving four resistors. The correct equivalent resistance was confirmed, while the initial calculations for current i1 and voltage across the 60-ohm resistor were incorrect. Using Kirchhoff's Voltage Law, the corrected current i1 was determined to be 3 A, and the voltage across the 60-ohm resistor was accurately calculated as 120 V. The importance of correctly identifying current through resistors in a circuit was emphasized.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Knowledge of series and parallel resistor combinations
  • Familiarity with current and voltage divider rules
  • Basic circuit analysis techniques
NEXT STEPS
  • Study Kirchhoff's Current Law (KCL) for current calculations
  • Learn about series and parallel resistor configurations in detail
  • Explore advanced circuit analysis techniques using Thevenin's and Norton's theorems
  • Practice solving circuit problems using simulation software like LTspice
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Electrical engineering students, circuit designers, and anyone involved in circuit analysis and design will benefit from this discussion.

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 through 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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