How Is Current Distributed in a Complex Circuit?

  • Thread starter Thread starter crhscoog
  • Start date Start date
  • Tags Tags
    Circuits
Click For Summary
SUMMARY

The discussion focuses on calculating current and power consumption in a complex circuit involving resistors of 20 ohms, 30 ohms, and 100 ohms. The user combined the 20 ohm and 30 ohm resistors to create a total resistance of 50 ohms, which, when placed in parallel with the 100 ohm resistor, resulted in a total resistance of approximately 33.33 ohms. Using Ohm's Law (V=IR), the current through the circuit was determined to be 0.277 A, leading to a total power consumption of 3.322 W. The user concluded that it would take approximately 50 minutes and 10 seconds to provide 10 kJ of electrical energy.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Knowledge of series and parallel resistor combinations
  • Familiarity with power calculations (P=RI^2)
  • Basic unit conversions (Joules to hours)
NEXT STEPS
  • Study advanced resistor network analysis techniques
  • Learn about Kirchhoff's laws for circuit analysis
  • Explore energy consumption calculations in electrical circuits
  • Investigate the implications of power factor in AC circuits
USEFUL FOR

Electrical engineering students, circuit designers, and anyone interested in understanding current flow and power consumption in complex electrical circuits.

crhscoog
Messages
16
Reaction score
0

Homework Statement


frqprob.jpg


The Attempt at a Solution



i. First I combined 20ohm and 30ohm to make 50ohms. Since this 50ohms and 100ohms is in parallel I have to use 1/Rtotal= 1/r1 + 1/r2... etc and got 33.333 ohms. I added this to 10ohms which ends up being 43.333 ohms. Then I used V=IR and got I=.277. So would .277 be the current going through 10ohm and half of that going through 100ohm and quarter of .277 going through 20ohm and 30ohm?

So from here I would get half of .277 as current for 100ohm and a fourth of .277 as current for both 20 and 30 ohm. and using P=RI^2 i would get a total of 2.925W for power consumption??

and with 2.925 I determine that it takes 3418.803419 hours to provide 10kJ of electrical energy??
 
Last edited:
Physics news on Phys.org
1. I = 0.277A right? Total power consumption assumes the total resistance, which is R = 43.3\Omega, right? So when I carried out the calculation, I ended up with P = 3.322 W

2. P = 3.322 W = 3.322 \frac{J}{s} Therefore, 10000 J = 3.322\frac{J}{s} \cdot x(s), right? So you will end up with x = \frac{10000 J}{3.322\frac{J}{s}} = 3010(s)

\frac{3010}{3600}h = 0.836 h = 50.16 min

So 50 minutes and 10 seconds is what it takes to provide 10kJ of energy.
 

Similar threads

  • · Replies 42 ·
2
Replies
42
Views
6K
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
3K