A 46 Mu F capacitor is connected to a signal generator at 8 volts maximum, 26Hz

Question

A 46 Mu F capacitor is connected to a signal generator at 8 volts maximum, 26Hz frequency. The phase angle between the current and voltage across the capacitor is (approximately): 60 degree 45 degree 0 degree 90 degree 30 degree 180 degree And the current the voltage. The magnitude of the maximum current in the circuit is mA If the generator frequency is increased the maximum current and the phase shift between the current and the voltage across the capacitor If another capacitor of the same value is added to the circuit in parallel to the original capacitor, the maximum current through the circuit will increase because the current in second resistor will be slightly out of phase with the current through the first capacitor decrease by x2 not change because the current in second capacitor will be pi/2 out of phase with the current through the first capacitor decrease because the current in second capacitor will be pi out of phase with the current through the first resistor increase by x2 not change because AC current will not branch into second capacitor

Explanation / Answer

Voltage lags Current by 900

Impedance(X)= 1/ (2*pi*f*C) = 133.07

Apply V= X*I

Imax= 8/133.07 = 0.06 A

phase shift between the voltage and current doesn't depend on the input frequency. So, its 900

If another capacitor attached in parallel X will be halfed.Because equivalent capapcitance would be 2*C and since X=1/2*pi*f*C , new X= 2*X Then by V= I*X;

Maximum current will be increased by x2

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