1.

Obtain the answers (a) to (b) in Exercise 15 if the circuit is connected to a 110 V, 12 kHz supply ? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.

Answer» Given, the ems value of voltage,
`V_("rms")=110 V`
The frequency of capacitor
`f=12kHz = 12000 Hz`.
Capacitance of conductor `C = 10^(-4)F`.
Tesistance `R = 40 Omega`
Capacitive Resistance
`X_(C )=(1)/(2pi fC)=(1)/(2xx3.14xx12000xx10^(-4))`
`= 0.133 Omega`
The rms value of current
`I_("rms")=(V_("rms"))/(sqrt(X_(C )^(2)+R^(2)))=(110)/(sqrt((40)^(2)+(0.133)^(2)))`
`= 2.75 A`.
The maximum value of current,
`I_(0)=sqrt(2) I_("rms")`
`= 1.414xx2.75`
`= 3.9 A`
Here, the value of `X_(C )` is very small, so term containing C is negligible.
`tan phi = (1)/(omega CR)`
`= (1)/(2xx3.14xx12000xx10^(-4)xx40)`
`= (1)/(96pi)`
It is very small.
In DC circuits, `omega = 0`
`X_(C )=(1)/(omega C)=oo`
So, it behaves like an open circuit.


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