Question

A 126 µF capacitor is connected to a source that operates at 120 Hz with a peak potential difference of 14 V. Find: a) the peak current;

b) the current when the potential difference has its peak value;

c) the current when the potential difference has one-half the (positive) peak value (there are two answers, list them in increasing order);

d) the instantaneous power delivered to the capacitor at 66 ms

Answer 1

a) The diagram and the waveform for this circuit would look like this:

$v_{c} = V_{c}sin\omega _{d}t$

where V_c is the amplitude of the AC voltage across the capacitor. And from the definition of capacitance, we say that the charge equals the capacitance times the voltage across it.

or,

q_c = Cv_c

and current is charge flowing per unit time.

and so, i_c = $\frac{dq_{c}}{dt}$ = $\frac{d(CV_{c}sin\omega_{d}t)}{dt}$ = $\omega_{d}(CV_{c}cos\omega_{d}t)$

which explains the waveform of the current above for why is must be out of phase (as cosine is out of phase with sine by 90 degrees).

and $\omega_{d} = 2\pi f_{d}$

therefore $\omega_{d} = 2\pi \times 120 = (240 \pi) rad/s$

and so $i_{c} = (240 \pi \times 126 \times 10^{-6} \times 14) cos(240\pi t)$

since cosine has maxima of +1, therefore the peak current would be:

$i_{c} = (240 \pi \times 126 \times 10^{-6} \times 14) 1$ = Ic = 1.33 A

b) As we saw from the graph and the relation above, the current would be out of phase with the voltage by 90 degrees. And so when the voltage (sine) will be maximum (peak value) which is at 90 degrees (since sin(90) = 1) , the current (cosine) should be cos(90) which is zero. And hence the current will be zero for peak voltage.

c) the peak value of voltage will be :

$v_{c} = V_{c}sin(90) = V_{c}$

so for v_c = V_c/2 , $sin(\omega_{d}t) = 1/2$

this will happen when $(\omega_{d}t) = 30 \degree or 150\degree$

and so the current for this value will be:

$i_{c} = (240 \pi \times 126 \times 10^{-6} \times 14) cos(30\degree)$ A or $i_{c} = (240 \pi \times 126 \times 10^{-6} \times 14) cos(150\degree)$ A

i_c = 1.15 A and -1.15 A (-1.15A and + 1.15A : increasing order)

d) at t = 66 ms, $\omega_dt = 240\pi \times 66 \times 10^{-3}$ = 49.76 degrees

for this value, the instantaneous voltage should be: $v_{c} = V_{c}sin(49.76\degree) = 0.763V_{c}$ = 10.68V

and the instantaneous current should be: $i_{c} = (240 \pi \times 126 \times 10^{-6} \times 14) cos(49.76\degree)$ = 0.86A

and Power is given by Voltage across a load times the current through it.

therefore the instantaneous power should P = 10.68 x 0.86 = 9.17 W