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I still need the voltage across the capacitor and resistor at each time. 3. An R

ID: 1499080 • Letter: I

Question

I still need the voltage across the capacitor and resistor at each time.

3. An RC circuit is constructed with a 1071 resistor and a 31,F capacitor connected to 12 V. a) What is the initial current in the circuit? b) After how long will the current drop to 1/2 its initial value? c) After how long will the current drop to 1/10 its initial value? 1/100? 1/1000? d) What is the voltage across the capacitor and the resistor at each time? Comment Expert Answer devshi answered this 26 minutes later 604 answers Was this answer helpful? Initial charge on capacitor o-C"V- 3 10-6*12 36 10A-6 C Intial current: 1.-VR = 12/1071-00112 A (Bo) Vlo =e^(-VT): 112 UT:0.693 t0.693 RC (since T time costant 0.693 1071 3-10-6 2.23 miliseconds. (c)For 1/10 of initial value of current: R*C)

Explanation / Answer

Hi,

In this case I will answer just the last part as the previous ones were already solved.

d) As you were told, the way to find the voltage through each element is:

For the voltage across the resistor

Vr = IR (in this case we assume that the material is ohmic)

For the voltage across the capacitor

Vc = V - Vr

The current at any moment in this circuit will be equal to:

I(t) = (V/R) exp(-t/(RC))

Assuming that the previous times are right, we introduce those times in the previous equation and we find the voltages.

For 1/2 (t = 2.23*10-3 s)

I = (12V/1071) exp[-2.23*10-3 s / (1071*3*10-6 F)] = 6.0*10-3 A

Vr = IR = (6.0*10-3 A)(1071) = 6.43 V

Vc = V - Vr = (12 - 6.43) V = 5.57 V

For 1/10 (t = 7.3*10-3 s)

I = (12V/1071) exp[-7.3*10-3 s / (1071*3*10-6 F)] = 1.2*10-3 A

Vr = IR = (1.2*10-3 A)(1071) = 1.29 V

Vc = V - Vr = (12 - 1.29) V = 10.71 V

For 1/100 (t = 14.8*10-3 s)

I = (12V/1071) exp[-14.8*10-3 s / (1071*3*10-6 F)] = 1.1*10-4 A

Vr = IR = (1.1*10-4 A)(1071) = 0.12 V

Vc = V - Vr = (12 - 1.29) V = 11.88 V

For 1/1000 (t = 22.19*10-3 s)

I = (12V/1071) exp[-22.19*10-3 s / (1071*3*10-6 F)] = 1.1*10-5 A

Vr = IR = (1.1*10-5 A)(1071) = 0.012 V

Vc = V - Vr = (12 - 1.29) V = 11.988 V

I hope it helps