A 8.0 µF capacitor is charged by a 11.0 V battery through a
resistance R. The...
A 8.0 µF capacitor is charged by a 11.0 V battery through a
resistance R. The capacitor reaches a potential difference of 4.00
V at a time 3.00 s after charging begins. Find R in kO.
I've been using
, but I'm not sure if that is right since I just CAN'T get the
right answer...I keep getting 370.7 kO (incorrect).
Solutions
Expert Solution
Concepts and reason
Use the concept of charging of the capacitor to solve this problem.
First calculate the time constant by using the concept of time constant of the capacitor.
Find the resistance of the circuit by using the equation for the voltage across the capacitor when it was charging.
Fundamentals
The expression for the time constant of the RC circuit is as follows:
τ=RC
Here,τis the time constant, R is the resistance of the circuit, and C is the capacitance.
The expression for the voltage across the capacitor when it was charging is as follows:
Vc=Vs(1−e−τt)
Here,Vcis the voltage across the capacitor,Vsis the supply voltage, and t is the elapsed time.
The time constant for the RC circuit is,
τ=RC
Substitute (8.0μF)for C in the above equation.
τ=(8.0μF(106μF1F))R=(8.0×10−6F)R
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