Physics Capacitors

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Capacitors

RC Circuit : Charging a capacitor

Let the capacitor be initially uncharged hence no potential drop across the capacitor. As soon as the circuit completes, the charge begins to flow.
Let 'q' be the charge on the capacitor at certain instance & i be the current in the circuit. Then,

`iR+q/C = V` & `i =(dq)/(dt)`

`R (dq)/(dt) = (CV-q)/C`

`int_0^q -(dq)/(CV-q) = int_0^t-(dt)/(CR)`

`q = q_0(1-e^(-t/(RC)))`

where, `q_0 = CV =` maximum amount of charge stored on the plates

`(dq)/(dt) = i = V/R e^(-t/(RC)) = V/R e^(-t/(tau))`

Once we know the charge on the capacitor we also can determine the voltage across the capacitor,

`V_c(t) = q(t)/C = epsilon(1 - e^(-t/(tau)))`

At that time, the voltage across the capacitor is equal to the applied voltage source and the charging
process effectively ends,

`V_c = (q(t=oo) )/(CV)= Q/C = epsilon`

For current a capacitor acts as :
Short-circuit just after closing the switch. ie all the potential will drop across the resistance. current maximum
Open circuit a long time after closing the switch. ie all the potential will drop across the capacitor, no current in the branch of capacitor.

Let the capacitor be initially uncharged hence no potential drop across the capacitor. As soon as the circuit completes, the charge begins to flow.
Let 'q' be the charge on the capacitor at certain instance & i be the current in the circuit. Then,

`iR+q/C = V` & `i =(dq)/(dt)`

`R (dq)/(dt) = (CV-q)/C`

`int_0^q -(dq)/(CV-q) = int_0^t-(dt)/(CR)`

`q = q_0(1-e^(-t/(RC)))`

where, `q_0 = CV =` maximum amount of charge stored on the plates

`(dq)/(dt) = i = V/R e^(-t/(RC)) = V/R e^(-t/(tau))`

Once we know the charge on the capacitor we also can determine the voltage across the capacitor,

`V_c(t) = q(t)/C = epsilon(1 - e^(-t/(tau)))`

At that time, the voltage across the capacitor is equal to the applied voltage source and the charging
process effectively ends,

`V_c = (q(t=oo) )/(CV)= Q/C = epsilon`

For current a capacitor acts as :
Short-circuit just after closing the switch. ie all the potential will drop across the resistance. current maximum
Open circuit a long time after closing the switch. ie all the potential will drop across the capacitor, no current in the branch of capacitor.

Discharging a Capacitor

Suppose initially the capacitor has been charged to some value Q. For t < 0 , the switch is open and the
potential difference across the capacitor is given by `V_c = Q/C`. On the other hand, the potential difference
across the resistor is zero because there is no current flow, that is, I= 0. Now suppose at t = 0 the switch
is closed (Figure). The capacitor will begin to discharge.

`iR - q/C = 0` & `i =(dq)/(dt)`

`R (dq)/(dt) = (CV-q)/C`

`int_(q_0)^q (dq)/q = int_0^t-(dt)/(CR)`

`q = q_0(e^(-t/(RC)))`

where, `q_0 = CV =` maximum amount of charge stored on the plates

`(dq)/(dt) = i = V_0/R e^(-t/(RC)) = V/R e^(-t/(tau))`

At t = 0 discharging current will be maximum then decay exponentially after long time will become zero

Suppose initially the capacitor has been charged to some value Q. For t < 0 , the switch is open and the
potential difference across the capacitor is given by `V_c = Q/C`. On the other hand, the potential difference
across the resistor is zero because there is no current flow, that is, I= 0. Now suppose at t = 0 the switch
is closed (Figure). The capacitor will begin to discharge.

`iR - q/C = 0` & `i =(dq)/(dt)`

`R (dq)/(dt) = (CV-q)/C`

`int_(q_0)^q (dq)/q = int_0^t-(dt)/(CR)`

`q = q_0(e^(-t/(RC)))`

where, `q_0 = CV =` maximum amount of charge stored on the plates

`(dq)/(dt) = i = V_0/R e^(-t/(RC)) = V/R e^(-t/(tau))`

At t = 0 discharging current will be maximum then decay exponentially after long time will become zero

Q 2108745608

Complete each sentence with a· correctly placed adverb and verb.

As long as I can remember, Sophie______ good at English.

(A)

has been always

(B)

always has been

(C)

has always been

(D)

been always had

Solution:

As long as I can remember, Sophie has always been good at English.

Correct Answer is `=>` (C) has always been

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