What is the time taken for an RC circuit to gain a charge equal to half the charge it would have at infinity?A steady voltage V is applied to a circuit of resistance R and capacitance C. Using...

What is the time taken for an RC circuit to gain a charge equal to half the charge it would have at infinity?

A steady voltage V is applied to a circuit of resistance R and capacitance C. Using Kirchoff’s law, it can be shown that the charge, q, in the circuit is given by the formula q= CV(1-e^-t/RC) where t is the time. Express the time t in terms of q, C, V and R. Hence find the time taken for the charge to reach half its terminal value (i.e. half of its value at infinite time).

 

Asked on by mbaya

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The steady voltage V is applied to a circuit consisting of a resistor with a resistance R and a capacitor with a capacitance C. The charge q in the circuit changes with time t and is given by the relation q = CV(1 - e^(-t/RC))

If the voltage is applied for an infinite duration of time, the charge in the circuit is given by CV*(1 - e^-inf.) = CV*(1 - 0) = CV

From q = CV(1 - e^(-t/RC))

=> q/CV = (1 - e^(-t/RC))

=> 1 - (q/CV) = e^(-t/RC)

take the natural log of both the sides

=> ln(1 - q/CV) = (-t/RC)*ln e

ln e = 1

=> ln(1 - q/CV) = -t/RC

=> t = -RC*ln(1 - q/CV)

For q = CV/2

t =   -RC*ln(1 - (CV/2)/CV)

t =   -RC*ln(1 - (1/2))

t =   -RC*ln(1/2)

t = RC*0.693

The RC circuit has half the charge it has at infinity after a duration of time equal to 0.693*RC

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