Question

A series circuit consists of R=8500 Ω and C=55 microF

(a) What is the value of the time constant?

(b) If the above circuit is connected to a 5V dc power supply, how long will it take for the voltage

across the capacitor to reach 50% of its maximum value?

(c) If the capacitor in the above circuit is fully charged and suddenly shorted, how long will it take

for the voltage across the capacitor to reach 20% its maximum value?

Answer 1

R=8500 Ω and C=55 microF

(a) Time constant= resistance* capacitance= R*C= 8500* 55/10^6= 0.4675s

(b) Using V= Vmax* e^(-t/RC)

or, 0.5Vmax= Vmax*e^(-t/RC)

or, 0.5= e^(-t/RC)

or, -0.693= -t/0.4675s

So, t= 0.324s

So, If the above circuit is connected to a 5V dc power supply,it will take 0.324s for the voltage

across the capacitor to reach 50% of its maximum value

(c) Now, using V= Vmax* e^(-t/RC),

or, 0.2Vmax= Vmax*e^(-t/RC)

or, 0.2= e^(-t/RC)

or, 1.6094= t/0.4675s

so, t= 0.7524s

So, If the capacitor in the above circuit is fully charged and suddenly shorted,  it will take  0.7524s

for the voltage across the capacitor to reach 20% its maximum value.

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