Finding the current in resistors

Click For Summary
SUMMARY

The discussion centers on calculating the current (i) through resistor R1 in a circuit with resistors R1 = 99Ω, R2 = R3 = 49.5Ω, R4 = 75.0Ω, and an ideal battery with an emf of 6.00 V. The user initially calculated the equivalent resistance (Req) of R2, R3, and R4 as 18.609Ω and applied the Loop Rule to derive the equation E - i(R1 + Req) = 0. However, the user miscalculated the current, arriving at 0.51016494 A instead of the correct value of 0.051 A due to a decimal placement error.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Familiarity with the Junction Rule in circuit analysis
  • Knowledge of the Loop Rule for voltage in circuits
  • Ability to calculate equivalent resistance in series and parallel circuits
NEXT STEPS
  • Review the application of the Loop Rule in circuit analysis
  • Practice calculating equivalent resistance for complex resistor networks
  • Explore advanced circuit analysis techniques using Kirchhoff's laws
  • Learn about error analysis in electrical calculations to avoid common mistakes
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone involved in circuit analysis or troubleshooting electrical circuits.

Oijl
Messages
102
Reaction score
0

Homework Statement


In the figure, R1 = 99 . R2 = R3 = 49.5 . R4 = 75.0 , and the ideal battery has emf script e = 6.00 V.

Re-GLORY.gif


What is i in R1?

Homework Equations


V=IR
Junction Rule
Loop Rule?
i = E / (R+r)


The Attempt at a Solution


The Loop Rule states that for any loop in a circuit, the sum of the voltages across the things in it will be zero. So by combining R2 R3 R4 into Req of resistance 18.609 ohms, I have a loop, so I add the voltages:

E - iR1 - iReq = 0
E = 6 as given in the problem. (Right? Maybe I'm wrong here.)
By the Junction Rule, the current i at R1 and Req would be the same.
So,
6 - i(R1 + Req) = 0
But this just gives me the equivilant resistance for all the resistors, and let's call that plain R.
So I have i = E/R, which is an equation I already had, so I felt good about coming across it via application of the two rules, but when I use this to find the current I don't get the right answer. My numerical answer to the question is 0.51016494 A.

I don't exactly know where I'm going wrong here...

Thank you for looking at this.
 
Physics news on Phys.org
I get the 18.6 ohms, too. Adding R1 in series, we have 117.6 ohms.
I = V/R = 6/117.6 = .051 A. Looks like you misplaced the decimal point. Quite forgivable at this time of night.
 
What is the right answer?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
5
Views
2K
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
10K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K