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Home/ Questions/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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lizawr
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lizawr
Asked: April 10, 20222022-04-10T08:34:19+00:00 2022-04-10T08:34:19+00:00In: Physics

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.

  • 26

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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.

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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...

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  

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  

Part C

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

Express your answer using two significant figures.

 
   
 
Q =     ?C  

Part D

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

Express your answer using two significant figures.

 
   
 
I =     mA  

Part E

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

Express your answer using two significant figures.

 
   
 
Q =     ?C  

Part F

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

Express your answer using two significant figures.

 
   
 
I =     mA

♦ Relevant knowledge
The term “capacitor” refers to a type of device that stores electricity in the form of an electrical field. The capacitor is composed of two conductors that are near but separated from each the other by an insulator or non-conductive zone. The insulator is also referred to as dielectric. Capacitors are mostly used for storage of energy.

A capacitor is composed of 2 plates, a positive and one that is negative. When electrical current moves throughout the electric circuit the capacitor absorbs energy on it’s positive plate. Equal amounts of charge is transferred to the negative plate at the same time and maintains equal charges on both plates. The capacitor is able to store the energy collected in the event that the electrical circuit was shut off.

capacitorresistor
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    tonytutoria
    2022-04-10T09:00:47+00:00Added an answer on April 10, 2022 at 9:00 am
    Concepts and Reason

    This problem can be solved by charging an RC circuit.

    The charge on a capacitor can initially be calculated by multiplying its capacitance and the voltage of a battery. Ohm’s law later allows you to calculate the current flowing through the circuit. The instantaneous charge of the capacitor can then be calculated by multiplying the instantaneous voltage with the capacitance.

    The ratio of the instantaneous resistance to the voltage can then be used to calculate the instantaneous charge. The instantaneous charge of the capacitor can also be calculated by multiplying the instantaneous voltage with the capacitance. The ratio of the instantaneous resistance and the voltage can be used to calculate the instantaneous charge.

    Fundamentals

    Below is the expression for the charge of the capacitor:

    Q=CV

    represents the voltage, Q is the charge and C is the capacitance.

    This is the expression of the current within the circuit:

    is the current and is the resistance in the circuit.

    Below is the expression for the instantaneous voltage:

    V=Vel/RC)

    is the instantaneous voltage, and represents time.

    This is the expression of the instantaneous capacitor charge:

    O=CV

    is the instantaneous cost.

    Below is the expression for the instantaneous current:

    is the instantaneous current.

    (A)

    This is the expression for the charge on a capacitor:

    Q=CV

    Substitute 2 μF to C and 9 V to V

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

    (B)

    This is the expression for current:

    Substitute for, for.

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

    (c)

    This is the expression for instantaneous voltage:

    V=Vel/RC)

    Substitute to ; to ; 2.0 uF
to ; 2.0 uF
and respectively for .

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

    Below is the expression for an instantaneous cost.

    O=CV

    Substitute 5.459 V to and 2 μF to C.

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

    (d)

    This is the expression for instantaneous current:

    Substitute 5.459 V to or 50 Ω to .

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

    (e)

    This is the expression for instantaneous voltage:

    Substitute to ; to ; 2.0 uF
to ; 2.0 uF
to 2.0 uF
; 2.0 uF
; 200 μs

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

    Below is the expression for an instantaneous cost.

    Substitute 1.218 V to or 2.0 μF to .

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

    (f)

    This is the expression for instantaneous current:

    Substitute 1.218 V
,

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

    Ans Part A

    The charge on a capacitor is 18 με

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