Question 1
A student is using a
power supply that produces a sinusoidal output. The meters on the supply show
that the output voltage V has a root-mean-square
(r.m.s.) value of 14 V with a frequency of 750 Hz.
The variation with
time t of the output voltage V
may be represented by the expression
V = V0 sin
ωt.
(a)
Determine the value of
(i)
V0,
[1]
(ii)
ω. [1]
(b)
A capacitor with a large capacitance is connected across the
terminals of the supply.
Suggest and explain
why this may lead to a large current from the supply. [3]
Reference: Past Exam Paper – November 2015 paper 43 Q7
Solution:
(a)
(i)
{Vrms
= V0 / √2
V0
= Vrms × √2}
V0 (= 14 √2 ) = 19.8 (20) V
(ii)
{ω = 2πf}
ω (= 2π ×
750) = 4700 rad s–1
(b)
{Capacitance
C = Q / V
Since the
capacitance is large, the amount of charge Q required will also be large.}
A large
amount of charge is required to charge the capacitor {of large capacitance}.
{The
output of the supply is sinusoidal, so the capacitor would charge and discharge
very rapidly (depending on the frequency). The high frequency gives rise to a
very short charging/discharging time.}
The capacitor
would charge and discharge rapidly.
Can you please explain in more detail how exactly does a large frequency give rise to short discharging/charging time? Thanks a lot!
ReplyDeletee.g. frequency = 750 Hz
DeleteThis means that in 1 second, the capacitor charges and discharges 750 times. So, the time for 1 charging/discharging should be very small in order for this process to occurs 750 times in 1 second.
In this question is the capacitor connected across (in parallel) to the supply (battery) or the output (say resistor)? A bit confused
ReplyDeleteit is in parallel to the supply
Delete