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The switch in the figure (Figure 1) has been in position a for a long time. It is changed to position b at t=0s. Par...

The switch in the figure (Figure 1) has been in position a for a long time. It is changed to position b at t=0s.

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Part A

What is the charge Q on the capacitor immediately after the switch is moved to position b?

Express your answer using two significant figures.

Q =   ?C  

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Part B

What is the current I through the resistor immediately after the switch is moved to position b?

Express your answer using two significant figures.

I =   mA  

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Part C

What is the charge Q on the capacitor at t=50?s?

Express your answer using two significant figures.

Q =   ?C  

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Part D

What is the current I through the resistor at t=50?s?

Express your answer using two significant figures.

I =   mA  

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Part E

What is the charge Q on the capacitor at t=200?s?

Express your answer using two significant figures.

Q =   ?C  

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Part F

What is the current I through the resistor at t=200?s?

Express your answer using two significant figures.

I =   mA
1 1
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Answer #1
Concepts and reason

The concept used to solve this problem is the charging of a RC circuit.

Initially, the charge on the capacitor can be calculated by multiplying the capacitance of the capacitor and the voltage of the battery. Later, the current flowing in the circuit can be calculated by using Ohm’s law. After that, the instantaneous charge on the capacitor can be calculated by determining the instantaneous voltage and then multiplying that with capacitance.

Then, the instantaneous current can be calculated by using the ratio of the instantaneous voltage to the resistance. Again, the instantaneous charge on the capacitor can be calculated by determining the instantaneous voltage and then multiplying it with capacitance. Finally, the instantaneous current can be calculated by using the ratio of the instantaneous voltage and to the resistance.

Fundamentals

The expression for the charge on the capacitor is given below:

Q=CV

Here, is the voltage, is the charge, and is the capacitance.

The expression of the current in the circuit is as follows:

Here, is the current, is the resistance connected in the circuit.

The expression of the instantaneous voltage is given below:

V=Vel/RC)

Here, is the instantaneous voltage and is time.

The expression of the instantaneous charge on the capacitor is as follows:

O=CV

Here, is the instantaneous charge.

The expression of the instantaneous current is given below:

Here, is the instantaneous current.

(A)

The expression for the charge on the capacitor is as follows:

Q=CV

Substitute 2.0 uF
for and for .

e-[(2-)(103}}ov)
= 18x10C
18 uc

(B)

The expression for the current is as follows:

Substitute for, for.

I=
(9.0 V)
(502)
= 0.18 A

(c)

The expression for the instantaneous voltage is as follows:

V=Vel/RC)

Substitute for , for , 2.0 uF
for , and for .

{sousy mo
V=(9.0V)|el(600](2018 months
= 5.459 V

The expression for the instantaneous charge is given below:

O=CV

Substitute 5.459 V
for and for .

Q=(5.459 V)|(2 uF)/ 1x10^F)]
21
= 10.918x10°C
*11uC

(d)

The expression for the instantaneous current is as follows:

Substitute 5.459 V
for and for .

1- (5.459 v)
(502)
=0.10918 A
= 109.17 mA
110 mA

(e)

The expression for the instantaneous voltage is as follows:

Substitute for , for , 2.0 uF
for and 200 μs
for .

(2013) $100
V=(9.0V)) el(900(2018)
= 1.218 V

The expression for the instantaneous charge is given below:

Substitute 1.218 V
for and 2.0 uF
for .

Q=(1218 V)|(20) LXION
*
1uF )
= 2.436x10c
= 2.4 uc

(f)

The expression for the instantaneous current is as follows:

Substitute, 1.218 V
for and for .

1-(1.218 V)
(502)
= 0.02436 A
24 mA

Ans: Part A

The charge on the capacitor is18 με
.

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