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Mathematical Equation for Capacitor Charging Voltage

Considering the circuit below to find out the charging current. When the switch is closed, the capacitor will start to charge.

capacitor%20charging

Applying Kirchhoff’s Voltage Law, we get

EvRvC=0(a)
or,EvC=vR=iCR=Rdqdt=Rddt(CvC)=RCddtvC
or,dtRC=dvCEvC
or,tRC+k1=lnEvC [Integratingboththesides]
or,tRC+k=lnEvC.b

Where, k is constant of integration. It can be calculated by putting initial time, t = 0 when capacitor voltage, vC=0V then, k=lnE

Substituting k=lnE on (b) we get,

tRC+lnE=ln(EvC)
or,tRC=ln(EvC)lnE
or,tRC=ln(EvCE)
or,etRC=EvCE
or,EetRC=EvC
or,vC=EEetRC
or,vC=E(1etRC)
vC=E(1etτ)[Where,τ=RC=Timeconstant]..c

After a several period of time constant, the capacitor will be as charged as the applied potential E. It is approximately 5 times of the time constant, τ to be fully charged. After 5τ{or >5τ} the capacitor voltage will be 20V as per the circuit shown above.


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