Correct, computer gets: 2.09E-02 s Hint: The time constant is the time in which
ID: 1493060 • Letter: C
Question
Correct, computer gets: 2.09E-02 s
Hint: The time constant is the time in which the charge on the capacitor has increased to a factor of (1-1/e) of its equilibrium value.
11. [2pt]
For the circuit in the previous problem, if the capacitor is initially uncharged and the switch is closed at time t = 0, what is the charge on the capacitor at time t = 2.59×10-2 s?
Answer: Last Answer: 100.84 C
Not yet correct, tries 1/20
12. [2pt]
For the circuit in the previous problem, what is the current through the circuit at t = 2.59×10-2 s after the switch is closed?
Answer: Last Answer: 3.94 A
Not yet correct, tries 2/20
13. [2pt]
For the circuit in the previous problem, what is the voltage across the capacitor at t = 2.59×10-2 s after the switch is closed?
Answer: Last Answer: 3.48 V
Not yet correct, tries 2/20
Explanation / Answer
Time constant of an RC circuit is = RC = 721 x 29 x 10-6 = 0.020909 seconds = 2.09 x 10-2 sec
11.) Charge on a charging capacitor is given by the equation
Q = CV (1 - e-t/RC )
so, at t= 2.59 x 10-2 = 0.0259 ,
Q = 29 x 10-6 x 12 x (1 - e-0.0259/0.020909 ) = 2.471633943 x 10-4 Coulombs
12.) Current through a charging capacitor is given by the equation I = (V/R) e-t/RC
therefore at t= 0.0259,
I = (12/721) e-0.0259/0.020909 = 4.822641243 x 10-3 Amperes
13.) Voltage across a charging capacitor is given by V = Vo (1 - e-t/RC )
where Vo is the maximum voltage, which is 12 here.
so, at t =0.0259, V = 12 (1 - e-0.0259/0.020909 ) = 8.522875664 Volts