The power output of an electrical supply is 2.4 kW at a potential difference (p.d.) of 240 V.

Question 15

The power output of an electrical supply is 2.4 kW at a potential difference (p.d.) of 240 V. The two wires between the supply and a kettle each have a resistance of 0.50 Ω, as shown.

What is the power supplied to the kettle and what is the p.d. across the kettle?

power / kW     p.d. / V

A         2.3                   230

B         2.3                   235

C         2.4                   230

D         2.4                   235

Reference: Past Exam Paper – June 2018 Paper 12 Q32

Solution:

Answer: A.

Total power supplied = Power supplied to kettle + Power dissipated in wires

The 2 wires and the kettle are connected in series. So, the same current I flows through them.

Power P = VI

At the supply, P = 2.4 kW and V = 240 V

Current I = P / V = 2400 / 240 = 10 A

Power dissipated in each wire = I²R = 10² × 0.5 = 50 W

Total power dissipated in both wires = 2 × 50 = 100 W = 0.1 kW

Total power supplied = Power supplied to kettle + Power dissipated in wires

Power supplied to kettle = Total power supplied – Power dissipated in wires

Power supplied to kettle = 2.4 – 0.1 = 2.3 kW

e.m.f. from supply = p.d. across wires + p.d. across kettle

V = IR

p.d. across 1 wire = IR = 10 × 0.5 = 5 V

p.d. across both wires = 2 × 5 = 10 V

e.m.f. from supply = p.d. across wires + p.d. across kettle

p.d. across kettle = 240 – 10 = 230 V

A car of mass 500 kg is at rest at point X on a slope, as shown. The car’s brakes are released and the car rolls down the slope

Question 12

A car of mass 500 kg is at rest at point X on a slope, as shown.

The car’s brakes are released and the car rolls down the slope with its engine switched off. At point Y the car has moved through a vertical height of 30 m and has a speed of 11 m s^-1.

What is the energy dissipated by frictional forces when the car moves from X to Y?

A 3.0 × 10⁴ J                        B 1.2 × 10⁵ J                        C 1.5 × 10⁵ J            D 1.8 × 10⁵ J

Reference: Past Exam Paper – November 2015 Paper 12 Q16 & June 2018 Paper 12 Q19

Solution:

Answer: B.

As the car rolls down the slope, its gravitational potential energy is being converted to kinetic energy AND work done against friction.

Loss in GPE = Gain in KE + Work down against friction

mgh = ½ mv² + Work down against friction

Work down against friction = mgh – ½ mv²

Work down against friction = (500×9.81×30) – (½ × 500 × 11²)

Work down against friction = 116 900 J = 1.2 × 10⁵ J

An ideal operational amplifier (op-amp) has infinite voltage gain and infinite slew rate.

Question 2

An ideal operational amplifier (op-amp) has infinite voltage gain and infinite slew rate.

(a) State what is meant by

(i) the voltage gain, [1]

(ii) infinite slew rate. [2]

(b) A non-inverting amplifier circuit incorporating an ideal op-amp is shown in Fig. 8.1.

Fig. 8.1

The supply to the op-amp is +9 V / −9 V.

The voltage gain of the amplifier circuit is 12.

Determine the resistance of resistor R. [2]

(c) For the circuit of Fig. 8.1, the variation with time t of the input potential VIN to the amplifier is shown in Fig. 8.2.

Fig. 8.2

On Fig. 8.3, show the variation with time t of the output potential VOUT for time t = 0 to time

Fig. 8.3

[4]

[Total: 9]

Reference: Past Exam Paper – June 2016 Paper 42 Q8

Solution:

(a)

(i)

The voltage gain is the ratio of the voltage output to the voltage input.

(ii)

An infinite slew rate means that the changes in the output voltage occur immediately when the input voltage changes.

(b)

{This is a non-inverting amplifier circuit.

Gain = 1 + R_F / R_IN }

12 = 1 + R / (1.5×10³)

R = 16.5 kΩ

(c)

straight line from (0,0) to (0.75t1, 9.0 V)

horizontal line from endpoint of straight line to t1

+9 V to –9 V (or v.v.) at t1

correct line to t2

{Gain = V_OUT / V_IN       giving V_OUT = Gain × V_IN = 12 × V_IN

The output voltage cannot be greater than the supply voltage of 9 V.

When V_OUT = 9 V, V_IN = V_OUT / Gain = 9 / 12 = 0.75 V

So, when V_IN = 0.75 V (in Fig 8.2), V_OUT = 9 V (in Fig 8.3).

V_IN = 0.75 V at t = 0.75t₁. So, in Fig 8.3, the line passes through (0.75t₁, 9.0 V)}

{The output voltage cannot be greater than 9.0 V. So, even if V_IN increases, V_OUT remains 9.0 V. So, from 0.75t₁ to t₁ (this is where V_IN is maximum), V_OUT is still 9.0 V. This is a horizontal line in the graph to t₁.}

{At t₁, V_IN changes from positive to negative. So, V_OUT also changes from positive to negative. V_IN is maximum (negative), causing V_OUT to be – 9.0 V.}

{Just as for the positive V_IN, V_OUT = – 9.0 V when V_IN = – 0.75 V. This occurs at time t = 0.25t₂ in Fig 8.2. Hence, in Fig 8.3, we have a horizontal line from (t₁, – 9.0 V) to (0.25t₂, – 9.0 V).}

{From t = 0.25t₂ to t₂, the graph is a straight line joining the points (0.25t₂, – 9.0 V) to (t₂, 0 V).}