The circuit in Fig. P6.9 is driven by an alternating EMF V = V o sin( t) a) The

Question

The circuit in Fig. P6.9 is driven by an alternating EMF V = Vosin(t)

a) The switch is initially put into position A. The frequency of the AC driving voltage is adjusted until the amplitude of the current delivered to the circuit is maximized. What is this frequency?

b) The frequency is left untouched and the switch is now placed in the open position. Use this frequency for the rest of the problem. Compute the phase angle between the current and the driving EMF. Does the current lead or lag the EMF ? For what value of resistance R is the amplitude of the current delivered in the open position half that delivered in position A ?

c) The switch is finally put into position B. What is the amplitude of the current delivered to the circuit?

What is the average power supplied by the EMF source ?

Explanation / Answer

a) net resistance = R +j(wL - 1/w2C)

so current is max when net resistance is least

therefore wL -1/w2C = 0

therefore w = 1 / sqrt(2LC)

so f = 1 / 2sqrt(2LC)

b)Now net resistance = R +j(wL - 1/wC)

put w = 1 / sqrt(2LC) so net resistance = R - [j *sqrt(L/2C)]

thereforee =   - tan-1 { [sqrt(L/2C)] / R }

it is negative so current lags

current will be half when net resistance will be twice of net resistance of case a i.e. R

i.e. R2 + L/2C = (2R)2

therefore R = sqrt(L/6C)

c) Now net resistance = j(wL - 1/wC) = j *sqrt(L/2C)

so current amplitude = V / sqrt(L/2C)

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