Charging/Discharging Capacitor & Magnetic Field

Click For Summary
SUMMARY

The discussion centers on calculating the current and charge of a capacitor in a circuit subjected to an increasing magnetic field of 5 T/s. The circuit consists of a resistor (R=25 ohms) and a capacitor (C=95 µF). At the moment the circuit is placed in the magnetic field, the current is determined to be 0.4 A, flowing in a counterclockwise direction. The final charge (Qf) on the capacitor is calculated to be 9.5e-5 C, and the challenge lies in determining the time required for the capacitor to charge to 25% of this final charge.

PREREQUISITES
  • Understanding of capacitor charging equations, specifically Q(t)=Qf(1-e^(-t/RC))
  • Knowledge of Ohm's Law (V=IR) and its application in circuits
  • Familiarity with the concept of magnetic fields and their effects on electric circuits
  • Basic proficiency in exponential functions and their properties
NEXT STEPS
  • Learn how to derive the time constant (τ = RC) for capacitor charging
  • Study the effects of changing magnetic fields on electric circuits, particularly Faraday's Law of Induction
  • Explore advanced capacitor charging scenarios involving different resistor and capacitor values
  • Investigate the relationship between current, voltage, and charge in RC circuits
USEFUL FOR

Students and educators in electrical engineering, physics enthusiasts, and anyone involved in circuit analysis and design, particularly those focusing on capacitors and magnetic field interactions.

longcatislong
Messages
12
Reaction score
0

Homework Statement


The magnetic field in the region shown below is increasing by 5 T/s. The circuit shown (with R=25 , C=95 µF, dimensions 1 m by 2 m) is inserted into the region as shown. The capacitor is initially uncharged.

See Diagram: http://www.webassign.net/userimages/23-31.gif?db=v4net&id=50305

a.) What is the current in the circuit at the moment the circuit is put in the B field?
.4 A
direction: counterclockwise

b) How long will it take the capacitor to charge to 25% of its final charge?
?

c) How much charge will the capacitor hold after a long time (assuming this magnetic field rate increase is maintained)?
?


Homework Equations



Q(t)=Qf(1-e^t/rc)

Q=VC



The Attempt at a Solution



I'm using Q(t)=Qf(1-e^t/rc) to find time.

First, I found Qf by using Q=VC. I know that V=IR so, Q=IRC, =(.4)(25)(9.5e-6)
Qf=9.5e-5


Then, I used Q(t)=Qf(1-e^t/rc)

.25=9.5e-5(1-e^-t/[(9.5e-5*25)]

I'm not getting a real or correct answer for this.


Your assistance, any help you could offer is much appreciated. THANK YOU!
 
Physics news on Phys.org
I think you numbers are correct but when the charge is 25% that means 25% x final charge.
I think you have just forgotten to multiply by the final charge??
 

Similar threads

Capacitor circuit time constant problem
  • · Replies 2 ·
Replies
2
Views
2K
TIme needed for a capacitor to reach a fraction of its final charge
  • · Replies 2 ·
Replies
2
Views
2K
Capacitor Questions Charging and Discharging
  • · Replies 4 ·
Replies
4
Views
4K
Finding Current in Resistors & Charge of Capacitor
  • · Replies 4 ·
Replies
4
Views
2K
How to find voltage across capacitor in RLC circuit?
  • · Replies 3 ·
Replies
3
Views
2K
Ranking charge stored on three capacitors by a battery
  • · Replies 6 ·
Replies
6
Views
3K
Maximum charge on the plates of a capacitor
  • · Replies 2 ·
Replies
2
Views
2K
What does "Fully Charged Capacitor" mean?
  • · Replies 6 ·
Replies
6
Views
3K
Potential difference in Capacitors
  • · Replies 3 ·
Replies
3
Views
2K
Why can we add impedances in series?
  • · Replies 4 ·
Replies
4
Views
2K