a)The current leads the driving emf by pi/4 deg, because the driving frequency is less than the resonance frequency.
b)The circuit is purely capacitive, hence the device connected is a capacitor.
c)The impedence Z of the circuit will be:
Z = e0/I ; from current eqn given, I = 10 A and e = 100 sqrt(2)
Z = 100 x sqrt(2)/10 = 100 x 1.414/10 = 14.14 Ohm
We know that,
cos(theta) = R/Z
R = Z cos(theta) = 14.14 x cos(pi/4) = 10 Ohm
Hence, R = 10 Ohm
d)we know that
Z = sqrt (Xc^2 + R^2)
Xc = sqrt (Z^2 - R^2 ) = sqrt (14.14^2 - 10^2) = 10
1/w C = 10
C = 1/10 w = 1/10 x 10 = 1/100 = 10^-2 F
Hence, C = 10^-2 F
The circuit shown below contains an AC generator which provides a source of sinusoidally varying emf...
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