• Physics 9702 Doubts | Help Page 138

Question 682: [Current of Electricity]

(a)

(i) On Fig.1, sketch I – V characteristic for a filament lamp.

(ii) Explain how resistance of the lamp may be calculated for any voltage from its I – V characteristic.

(b) Two identical filament lamps are connected first in series, and then in parallel, to a 12 V power supply that has negligible internal resistance. The circuits are shown in Fig.2 and Fig.3 respectively.

(i) State and explain why resistance of each lamp when they are connected in series is different from the resistance of each lamp when they are connected in parallel.

(ii) Each lamp is marked with a rating ‘12 V, 50 W’. Calculate total resistance of the circuit for the two lamps connected such that each lamp uses this power.

Reference: Past Exam Paper – June 2011 Paper 23 Q5

Solution 682:

(a)

(i) The curve starts from (0,0) and is a smooth curve in correct direction.

Curve correct for end section never horizontal

(ii) Resistance R = V / I, hence we take the co-ordinates of V and I at a point from graph and calculate the ratio V / I.

(b)

(i) Each lamp connected in parallel has a greater p.d. / greater current across it {since the overall resistance in the circuit is less, the current is higher}. So, the lamp gets hotter. {The resistance of a lamp is not constant. As the current passes through it, the lamp gets hotter and thus, its resistance increases. This is indicated in the graph drawn in part (a). As the resistance of each of the lamps become greater, the combined resistance of the parallel combination increases.} Thus, the resistance of the lamps in parallel become greater.

(ii)

Power P = V² / R        or Power P = VI and V = IR

Resistance R {= V² / P = 12² / 50} = 144 / 50 = 2.88 Ω for each lamp

{For parallel combination, total resistance = (1/2.88 + 1/2.88)^-1 = 1.44 Ω}

Total Resistance = 1.44 Ω

Question 683: [Measurement + Resistance]

Resistance R of a uniform metal wire is measured for different lengths l of the wire. Variation with l of R is shown in Fig.1.

(a) Points shown in Fig.1 do not lie on the best-fit line. Suggest a reason for this.

(b) Determine gradient of the line shown in Fig.1.

(c) Cross-sectional area of the wire is 0.12 mm².

Use answer in (b) to determine resistivity of the metal of the wire.

                                                                                    

(d) Resistance R of different wires is measured. Wires are of the same metal and same length but have different cross-sectional areas A.

On Fig.2, sketch a graph to show variation with A of R.

Reference: Past Exam Paper – November 2014 Paper 21 Q3

Solution 683:

(a) The scatter of the points is due random error (in the measurements) of the length OR resistance

{Systematic error is associated with the apparatus and would produce a similar error for all the measurements.}

(b) {Consider points (0.8, 3.6) and (0.4, 1.9)}

Gradient = (3.6 – 1.9) / (0.8 – 0.4) = 4.25

(c)

Resistance R = ρl / A

{Resistivity ρ = RA / l = (R/l) A}

Resistivity ρ = gradient × area = 4.25 × (0.12 × 10^–6) = 5.1(0) × 10^–7 Ω m

                                                                                    

(d) The graph should show the resistance decreasing with increasing area.

Correct shape with curve being asymptote to both axes

{R = ρl/A. R is proportional to 1/A.}

Question 684: [Kinematics]

Variation with time t of velocity v of a car is shown in Fig.1.

At time t = 0, driver sees an obstacle in the road. A short time later, driver applies the brakes. The car travels in two stages, as shown in Fig.1.

(a) Use Fig.1 to describe the velocity of the car in

1. stage 1,

2. stage 2.

(b)

(i) Calculate distance travelled by the car from t = 0 to t = 3.5 s.

(ii) Car has a total mass of 1250 kg. Determine total resistive force acting on the car in stage 2.

(c) For safety reasons drivers are asked to travel at lower speeds. For each stage, describe and explain effect on the distance travelled for the same car and driver travelling at half the initial speed shown in Fig.1.

(i) stage 1:

(ii) stage 2:

Reference: Past Exam Paper – November 2011 Paper 23 Q2

Solution 684:

(a)

1. Constant velocity / speed

2.

EITHER constant / uniform decrease (in velocity/speed)

(b)

(i)

The distance travelled is given by the area under graph for both stages.

Stage 1: {area of rectangle} distance (18 × 0.65) = 11.7 (m)

Stage 2: {area of triangle} distance = (½ × 18 × [3.5 – 0.65]) = 25.7 (m)

Total distance {= 11.7 + 25.7} = 37.(4) m

(ii)

EITHER

F = ma

Acceleration a = (18 – 0) / (3.5 – 0.65)

Force F = 1250 × 6.3 =7900 N

OR

E_K = ½ mv² = ½ × 1250 × (18)²

{Force × distance = work done = E_K.             F × 25.7 = E_K}

F = ½ × 1250 × (18)² / 25.7 = 7900 N

OR

Initial momentum = 1250 × 18

F = change in momentum / time taken = (1250 × 18) / 2.85 = 7900

(c)

(i)

{Here, the speed is constant, that is, the acceleration is zero. Speed = distance / time.

Distance = speed × time. If speed is halved, the distance travelled is also halved.}

EITHER half / less distance as speed is half / less

OR half distance as the time is the same

{By moving at a lower speed, the reaction time of the driver may be faster. That is the time is shorter.}

OR sensible discussion of reaction time

(ii)

{v² = u² + 2as. Here the initial speed u is now halved. Final speed is zero. The acceleration / deceleration is constant. Here, the time to stop the car will also be less since the speed is lower.

When initial speed is u, Distance s₁ = u² / 2a

When initial speed is u/2, Distance s₂ = (u/2)² / 2a = u² / 8a = s₁/4}

EITHER same acceleration and s = v² / 2a

OR v² (for this) has now become ¼ v²

So, the distance travelled will be ¼ of the distance travelled {when the speed is not halved}

Question 685: [Current of Electricity]

Fig.1 shows a 12 V power supply with negligible internal resistance connected to uniform metal wire AB.

Wire has length 1.00 m and resistance 10 Ω. Two resistors of resistance 4.0 Ω and 2.0 Ω are connected in series across the wire.

Currents І₁, І₂ and І₃ in the circuit are as shown in Fig.1.

(a)

(i) Use Kirchhoff’s first law to state relationship between І₁, І₂ and І₃.

(ii) Calculate I₁.

(iii) Calculate the ratio x, where

x = power in metal wire / power in series resistors:

           

(b) Calculate potential difference (p.d.) between the points C and D, as shown in Fig.1. Distance AC is 40 cm and D is the point between the two series resistors.

Reference: Past Exam Paper – November 2012 Paper 22 Q5 & Specimen 2016 Paper 2 Q5

Solution 685:

(a)

(i) I₁ = I₂ + I₃

(ii)

I₁ is the total current in the circuit.

I = V / R                                              or I₂ = 12/10 = 1.2A

R = [1/(2+4) + 1/10]^-1 = 3.75Ω           or I₃ = 12/6 = 2.0A

I₁ = 12/3.75 = 3.2A                             or I₁ = 1.2 + 2.0 = 3.2A

(iii):

Power = VI                 OR I²R                        OR V²/R

x = (I₂²R_w) / (I₃R_s)       OR (VI₂) / (VI₃)          OR (V²/R_w) / (V²/R_s)

(choosing P=VI)

x = (12 × 1.2) / (12 × 2.0) = 0.60

           

(b) {p.d. V is proportional to resistance R [V = IR] and the resistance R of a wire is proportional to the length of the wire, L [R = ρL/A].

So, V is proportional to L. The wire is in parallel with the power supply, so the p.d. across AB (total length of the wire) is the same (= 12V) as the e.m.f. from the power supply. When the total length L of the wire is taken in account [point C is at point B], the p.d. across AC is 12V. As the length AC is decreased, the p.d. across AC will also decrease.

Now, point A and B are fixed and are connected to the terminals of the supply. The positive terminal of the supply and point A are at a potential of 12V. The negative terminal of the supply and point B are at a potential of 0V.

But, the p.d. V across AC is a potential difference – that is, a difference in potential.

When point C is 40cm from point A, p.d. across AC is 12 × (0.4/1) = 4.8V. This is a (p.d.) difference in potential. Since point A is at a potential of 12V, and the difference in potential between AC is 4.8V, point C will be at a potential of 12 – 4.8 = 7.2V. Point B is at 7.2V and point C is at 0V, so the p.d. across BC = 7.2V}

Since wire is 1m, p.d. across AC = 12 × (0.4/1)V and across BC = 12 – (12 × 0.4/1)V. 

p.d. BC = 12 – (12 × 0.4) = 7.2V       or p.d. AC = 4.8V

{Similarly, the total p.d. across resistors in series is the sum of p.d. across each resistor. So, across BD, p.d. = (12 × 2/6) or (12 – (12 × 4/6)) = 4.0V and p.d. across AD = (12 × 4/6) = 8.0V. This is the potential divider equation. Point D is at a potential of 4.0V}  

p.d. BD = 12 – (12 × 4/6) = 4.0V       or p.d. AD = 8.0V

p.d. between points C and D is the difference in potential.

p.d. = 7.2 – 4.0 = 3.2V                       or p.d. = 8.0 – 4.8 = 3.2V

{Question:

Why the pd between C and D has to be: pd of AD - pd of AC or pd of BC - pd of BD I just don't understand why. The current is in the same directions especially in BC and BD. Why are we subtracting them?

Explanation:

To find the p.d. between C and D, we need to find the potential at C and that at D and then take the difference.

Facts:

(1) Current flows from the +ve terminal of the supply.

(2) The –ve terminal of the supply is taken to be 0V and the +ve terminal is taken to be +12V. Any point connected directly to the +ve terminal (example: point A) is also at 12V and any point connected directly to the –ve terminal (example: point B) is at 0V. So, as we move from A to B, the potential decreases.

Potential DIFFERENCE (p.d.) between A and C = Potential at A – Potential at C

p.d. between A and C can be obtained from Ohm’s law: V = I₂R. The resistance of a wire is proportional to its length (R = ρL / A). 100cm corresponds to a resistance of 10Ω. 40cm corresponds to a resistance of (0.4 / 1) × 10 = 4Ω.

p.d. between A and C = I₂R = 1.2 × 4 = 4.8V

Potential at C = Potential at A – p.d.between A and C = 12 – 4.8 = 7.2V

p.d. between A and D = Potential at A – Potential at D

p.d. between A and D = p.d. across the 4.0Ω resistor = I₃ (4.0) = 2.0 × 4.0 = 8.0V

Potential at D = 12 – 8.0 = 4.0V

Potential difference between C and D = 7.2 – 4.0 = 3.2V}

Question 686: [Vectors]

Vector quantity V is resolved into two perpendicular components X and Y. Angle between V and component X is θ.

Angle between component X and the vector V is increased from 0° to 90°.

How do magnitudes of X and Y change as the angle θ is increased in this way?

X                     Y

A         increase           increase

B         increase           decrease

C         decrease          increase

D         decrease          decrease

Reference: Past Exam Paper – June 2010 Paper 11 Q5 & Paper 12 Q6 & Paper 13 Q2

Solution 686:

Answer: C.

When the angle θ is zero, the vector X would be the exactly same as vector V. In this situation, vector V does not have any component along Y. So here, vector Y would be zero. Vector X would have the maximum value here.

As the angle θ is increased, the magnitude of vector V is shared between a component along X and another component along Y. So, as the angle θ increases, the magnitude of X decreases while that of Y increases.

When the angle is 90°, the vector Y would be exactly the same as vector V. In this situation, vector V does not have any component along X. So, here, vector Y would be zero.