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.

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

The diameter of the cross-section of a long solenoid is 3.2 cm, as shown in Fig. 9.1.

Question 1

(a) State Faraday’s law of electromagnetic induction. [2]

(b) The diameter of the cross-section of a long solenoid is 3.2 cm, as shown in Fig. 9.1.

Fig. 9.1

A coil C, with 85 turns of wire, is wound tightly around the centre region of the solenoid.

The magnetic flux density B, in tesla, at the centre of the solenoid is given by the expression

B = π×10^-3 × I

where I is the current in the solenoid in ampere.

Show that, for a current I of 2.8 A in the solenoid, the magnetic flux linkage of the coil C

is 6.0 × 10^-4 Wb. [1]

(c) The current I in the solenoid in (b) is reversed in 0.30 s.

Calculate the mean e.m.f. induced in coil C. [2]

(d) The current I in the solenoid in (b) is now varied with time t as shown in Fig. 9.2.

Fig. 9.2

Use your answer to (c) to show, on Fig. 9.3, the variation with time t of the e.m.f. E induced in coil C.

Fig. 9.3

[4]

[Total: 9]

Reference: Past Exam Paper – November 2016 Paper 41 & 43 Q9

Solution:

(a)

Faraday’s law of electromagnetic induction states that the (induced) e.m.f. is proportional to the rate of change of (magnetic) flux (linkage).

(b)

{Area A = π × r²}

flux linkage = BAN

flux linkage = π × 10^-3 × 2.8 × π × (1.6 × 10^-2)² × 85 = 6.0 × 10^-4 Wb

(c)

e.m.f. = ΔNΦ / Δt

e.m.f. = (6.0 × 10^-4 × 2) / 0.30

e.m.f. = 4.0 mV

(d)

sketch:

{The e.m.f. induced is proportional to the rate of change of flux. If the current is constant (not changing), there is no change in flux. So, no e.m.f. is induced (zero).}

E = 0 for t = 0 → 0.3 s, 0.6 s → 1.0 s, 1.6 s → 2.0 s

{As calculated from part(c), when the current is reversed, the e.m.f. induced = 4 mV}

E = 4 mV for t = 0.3 s → 0.6 s (either polarity)

{From t = 1.0 s → 1.6, the e.m.f. induced is 2 mV (half the previous value) as the time 2 times longer. (e.m.f. = ΔNΦ / Δt).}

E = 2 mV for t = 1.0 s → 1.6 s with opposite polarity