9702/May Jun/11/2018/Q34

In the circuit shown, the batteries have negligible internal resistance.

What are the values of the currents I1, I2 and I3?

Solution:

Answer: C

Reference: PYQ - May/Jun 2018 Paper 11 Q34

9702/Oct Nov/13/2013/Q34

An electrical device of fixed resistance 20 Ω is connected in series with a variable resistor and a
battery of electromotive force (e.m.f.) 16 V and negligible internal resistance.

What is the resistance of the variable resistor when the power dissipated in the electrical device is 4.0W?

A 16 Ω 

B 36 Ω 

C 44 Ω 

D 60 Ω

Solution:

Answer: A

Power dissipated, P = I²R

For electrical device,

4.0 = I² (20)

Current I in circuit = 0.45A

Since this is a series circuit, the same current flows through the variable resistor.

Ohm’s law: V = IR

p.d. across electrical device = 0.45 (20) = 9.0V

Let the resistance of the variable resistor = R

From Kirchhoff’s second law, the sum of p.d. in a loop is equal to the e.m.f. in the circuit.

16 = 9.0 + 0.45R

Resistance R = 16Ω

Reference: PYQ - Oct/Nov 2013 Paper 13 Q34

9702/May Jun/11/2011/Q34

The resistance of a metal cube is measured by placing it between two parallel plates, as shown.

The cube has volume V and is made of a material with resistivity ρ. The connections to the cube
have negligible resistance.

Which expression gives the electrical resistance of the metal cube between X and Y?

Solution:
Answer: C

Resistance R of a wire = ρL / A

Where L is the length of the wire and A is the cross-sectional area of the wire.

For a cube, all the sides are of equal lengths. Let the length be L/

Volume V = L³

Length = L and cross-sectional area (in contact with the plates) = L²

Resistance R = ρL / L² = ρ / L

But since V = L³, length L = V^1/3

Resistance R = ρ / V^1/3

Alternatively, if we consider the units of the quantities involved, only choice C gives the unit of resistance (Ω).

Unit of resistivity = Ωm

Unit of volume V = m³

Unit of ρ / V^1/3 = [Ωm] / [m³]^1/3 = [Ωm] / m = Ω 

Reference: PYQ - May/Jun 2011 Paper 11 Q34

9702/May Jun/12/2011/Q34

A source of electromotive force (e.m.f.) E has a constant internal resistance r and is connected to
an external variable resistor of resistance R.

As R is increased from a value below r to a value above r, which statement is correct?

A The terminal potential difference remains constant.
B The current in the circuit increases.
C The e.m.f. of the source increases.
D The largest output power is obtained when R reaches r.

Solution:
Answer: D

Since the battery has some internal resistance, the terminal potential difference of the battery (p.d. across its terminals – this is the p.d. available to the rest of the circuit) is less than the e.m.f E.

The voltage lost in the batter due to its internal resistance = Ir where I is the current in the circuit.
Current I = E / (R + r)
As R increases, the current I decreases. [B is incorrect] 

Terminal pd, V = E – (Ir) 
Since current I changes with the resistance R, the terminal p.d. does not remain constant as R is being changed. [A is incorrect]

The e.m.f. of the source is constant, it does not increase. [C is incorrect]

Output power in the load, P = I²R = E²R / (R + r)² 
E and r are kept constant while R is being varied. The largest output power is obtained when R reaches r. 

This can be proved by differentiating P (in the above equation) with respect to R and then equating to zero.
dP / dR = [E²(R + r)² – 2E²R(R + r)] / (R + r)⁴ 

For maximum power P, dP/dR = 0
E²(R + r)² – 2E²R(R + r) = 0               {divide by E²(R + r) on both sides,}
(R + r) – 2R = 0
R = r

Reference: PYQ - May/Jun 2011 Paper 12 Q34