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A capacitor is discharged through a resistance. the time constant for the circuit is ?, then the value of time constant for power diddipated through the resistance will be ? what does time constant refers here?

A capacitor is discharged through a resistance. the time constant for the circuit is  ?, then the value of time constant for power diddipated through the resistance will be ?


 


 


what  does time constant refers here?


Grade:12

1 Answers

Aman Bansal
592 Points
12 years ago

Dear Pooja,

 V_{in}(s) = V\frac{1}{s}
 V_C(s) = V\frac{1}{1 + sRC}\frac{1}{s}

and

 V_R(s) = V\frac{sRC}{1 + sRC}\frac{1}{s} .
Capacitor voltage step-response.
Resistor voltage step-response.

Partial fractions expansions and the inverse Laplace transform yield:

 \,\!V_C(t) = V\left(1 - e^{-t/RC}\right)
 \,\!V_R(t) = Ve^{-t/RC} .

These equations are for calculating the voltage across the capacitor and resistor respectively while the capacitor is charging; for discharging, the equations are vice-versa. These equations can be rewritten in terms of charge and current using the relationships C=Q/V and V=IR (see Ohms law).

Thus, the voltage across the capacitor tends towards V as time passes, while the voltage across the resistor tends towards 0, as shown in the figures. This is in keeping with the intuitive point that the capacitor will be charging from the supply voltage as time passes, and will eventually be fully charged.

These equations show that a series RC circuit has a time constant, usually denoted τ = RC being the time it takes the voltage across the component to either rise (across C) or fall (across R) to within 1 / e of its final value. That is, τ is the time it takes VC to reach V(1 − 1 / e) and VR to reach V(1 / e).

The rate of change is a fractional \left(1 - \frac{1}{e}\right) per τ. Thus, in going from t = Nτ to t = (N + 1)τ, the voltage will have moved about 63.2 % of the way from its level at t = Nτ toward its final value. So C will be charged to about 63.2 % after τ, and essentially fully charged (99.3 %) after about . When the voltage source is replaced with a short-circuit, with C fully charged, the voltage across C drops exponentially with t from V towards 0. C will be discharged to about 36.8 % after τ, and essentially fully discharged (0.7 %) after about . Note that the current, I, in the circuit behaves as the voltage across R does, viaOhms Law.

Best Of luck

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Thanks

Aman Bansal

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