Question 1
Two capacitors P and
Q, each of capacitance C, are connected in
series with a battery of e.m.f. 9.0 V, as shown in Fig. 6.1.
Fig. 6.1
A switch S is used to
connect either a third capacitor T, also of capacitance C,
or a resistor R, in parallel with capacitor P.
(a)
Switch S is in position X.
Calculate
(i)
the combined capacitance, in terms of C,
of the three capacitors, [2]
(ii)
the potential difference across capacitor Q. Explain your working.
[2]
(b)
Switch S is now moved to position Y.
State what happens to
the potential difference across capacitor P and across capacitor Q. [4]
[Total: 8]
Reference: Past Exam Paper – November 2017 Paper 41 Q6
Solution:
(a)
(i)
{When switch S is in position X, capacitor T is
in parallel to capacitor P.
For parallel capacitors: overall capacitance = C
+ C = 2C
This combination of capacitors is in series
with capacitor Q.
Overall capacitance:}
1 / T = 1 / (2C) + 1 / C
T = ⅔C or 0.67C
(ii)
The same
charge is stored on capacitor Q as on the combination.
So, the p.d.
across Q is 6.0 V.
{Capacitor Q is in series with the combination
of the two capacitors P and T. So, the same current flows through capacitor Q
and the combination (though the current would split at the junction).
Current is the flow of charge.
So, the charge stored on capacitor Q is the
same as the charge in the combination.}
{Let V1 be the p.d. across capacitor
Q and V2 be the p.d. across the combination.
But V =
Q / C,
V1
= Q / C
e.m.f.
= total charge / total capacitance = Q / (2C/3) = 9.0 V
Q / (2C/3)
= 9.0 V
3/2 (Q/C)
= 9.0
Q/C = 9.0 × 2 / 3 = 6.0 V
So, V1 = Q/C = 6.0 V}
(b)
{When the switch S is moved to position Y,
capacitor P is now in parallel to the resistor R. This causes the capacitor P
to discharge through the resistor R whilst at the same time, capacitor Q would
charge.
So, the p.d. across P decreases from 3 V to 0 V
(when it is completely discharged) while capacitor Q charges up until the p.d.
across it is equal to the e.m.f. of the battery}
Capacitor P: p.d. will decrease (from 3.0 V) to
zero
Capacitor Q: p.d. will increase (from 6.0 V) to
9.0 V
IS IT SO THAT EVERYTIME THE CAPACITOR HAS TO BE CHARGED OR DISCHARGED THE RESISTOR WILL BE IN PARALLEL (AND NOT IN SERIES)?. PLZ EXPLAIN!!
ReplyDeleteby connecting in parallel, there would be the same p.d. across both.
Deleteif connected in series, some of p.d. would be across the resistor and some across the capacitor
can you further explain b? Why would the p.d. of the capacitor Q increase to 9V? Given that 3V is being discharged as heat in the resistor (is this correct thinking?), shouldn't Q remain at 6V? Or is it that the current simply doesn't flow through capacitance P, instead routing through resistor P, but if that is the case, why would that be?
ReplyDeleteI don't understand the theory of how capacitors and resistors work together in a circuit.
The p.d. across the resistor is now due to the capacitor P which acts as a source of e.m.f. for the resistor. Since energy is supplied to the resistor, the capacitor P discharges and the p.d. across it decreases.
DeleteNow, from Kirchhoff’s law, the sum of p.d. between P and Q should be equal to the e.m.f. Since the p.d. across P is decreasing, the p.d. across Q should increase to that the sum is still 9 V.
Thank you very much . It is really helpful . Please keep uploading P5 of physics also .
ReplyDeleteI have a doubt in your answer.
ReplyDeleteCapacitor Q basically became a battery after charging for a certain time right? Acting as a new source for the system?
Then why don't charges from Q flow to both capacitor P and resistor Q but instead P only discharge through resistor R which is parallel?
Q is being charged by the battery. Current cannot flow in opposite directions in the same component (here, capacitor Q). So, the current from the battery flows through P while charges from capacitor P flows through the resistor. This causes the p.d. across P to decrease and hence the p.d. across Q increases.
DeleteWhy doesn't capacitor P get charged too alongside capacitor Q while being discharged, since it's still connected to the battery? Really confused.
ReplyDeleteCapacitor P is discharging through the resistor. It cannot charge and discharge at the same time. Even if it is connected to the battery, most of the e.m.f. of the battery appears as the p.d. across Q, and not P.
Delete