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 Question 458: [Current of Electricity > Potential difference]
A 20 V d.c. supply is connected to circuit consisting of five resistors L, M, N, P and Q.

There is potential drop of 7 V across L and a further 4 V potential drop across N.
What are potential drops across M, P and Q?

Reference: Past Exam Paper – November 2013 Paper 13 Q38



Solution 458:

Answer: C.

This question can be easily tackled by considering the different loops present in the circuit and apply Kirchhoff’s laws to them.

Consider the loop: ‘+’ terminal supply – resistor L – resistor M – ‘-’ terminal supply

From Kirchhoff’s law, the sum of p.d. across any loop should be equal to the e.m.f. Analysis of the loop shows that a 7 V drop across resistor L must mean a 13V drop across M (to obtain a total 20V across L and M).

Notice that the direction of potential drop is also of significance.

Consider resistor L for example. The direction of potential drop is from left to right. Current flows from the ‘+’ terminal of the supply, so the junction on the left of resistor L should be at a higher potential (which is equal to 20V since there is no component between it and the ‘+’ terminal of the supply).

So, the direction of potential rise is from a greater value of potential to a small value of potential. Additionally, current flows from a greater potential to a smaller potential (as in the case of the ‘+’ terminal of the supply).

The junction between L and M is at a potential of 13V (since the right junction to which M is connected is at 0V as it is connected to the ‘-’ terminal of the supply).

There is a potential drop of 4V downwards across N, so current flows downwards. This means that the potential at the upper junction (between L and M) is greater than the lower junction (between P and Q) and the difference in potential is 4V.

Thus, lower junction (between P and Q) is at a potential of (13V – 4V =) 9V

So, the potential drop across resistor Q is 9V (since the right junction to which Q is connected is at 0V as it is connected to the ‘-’ terminal of the supply).

Finally, consider resistor P. Its terminal is at a potential of 20V and its right terminal is at a potential of 9V. Potential drop = 20 – 9 = 11V.

Important notice:

When considering a loop, it should start from one terminal of the supply and end at the other terminal. Only then will Kirchhoff’s law apply. For example, ‘+’ terminal supply – L – N – P – ‘+’ terminal supply is not a correct loop. You may notice that the sum of p.d. is not equal to the e.m.f. Additionally the flow of current is wrong.

Question 459: [Current of Electricity]
Diagram shows a length of track from model railway connected to a battery, a resistor and a relay coil.

With no train present, there is a current in relay coil which operates a switch to turn on a light.
When a train occupies the section of track, most of the current flows through wheels and axles of the train in preference to the relay coil. The switch in the relay turns off the light.
Why is a resistor placed between the battery and track?
A to limit the heating of the wheels of the train
B to limit the energy lost in the relay coil when a train is present
C to prevent a short circuit of the battery when a train is present
D to protect the relay when a train is present

Reference: Past Exam Paper – June 2013 Paper 13 Q35



Solution 459:

Answer: C.

When the train is present, it is stated that most of the current flows through the wheels and axles of the train in preference to the relay coil. So, there is little current in the relay coil. Thus, the energy lost in the relay coil when a train is present is already limited. The relay is also already protected. [B and D are incorrect]

Ohm’s law: V = IR giving current I = V / R

The resistor actually limits the current flowing in the circuit when the train is present. This prevents a short circuit of the battery when a train is present.

The heating of the wheels due to the current is small compared to that due to its motion.

Question 460: [Forces > Moment]

Rigid L-shaped lever arm is pivoted at point P.

Three forces act on the lever arm, as shown in diagram.

What is the magnitude of resultant moment of these forces about point P?

A 15 N m                    B 20 N m                    C 35 N m                    D 75 N m

Reference: Past Exam Paper – November 2010 Paper 11 Q13

Solution 460:

Answer: A.

We need to consider which forces will cause a clockwise moment and which cause an anticlockwise moment, that is consider how the lever would rotate about P due to each force.

Clockwise moment is caused by the 5N upward and 10N horizontal forces.

Anticlockwise moment is caused by the 15N downward force.

As for the distances, we need to consider the perpendicular distances from the line of action of the forces from point P.

Perpendicular distance of 5N force: (3m + 1m) – 2m = 2m

Perpendicular distance of 10N force: 2m

Perpendicular distance of 15N force: 3m

Magnitude of resultant moment = Anticlockwise moment – Clockwise moment

Magnitude of resultant moment = (15 × 3) – [(10 × 2) + (5 × 2)] = 15 N m

Question 461: [Electromagnetism > Hall probe]

Hall probe is placed distance d from long straight current-carrying wire, as illustrated. 

Direct current in wire is 4.0 A. Line XY is normal to wire.

Hall probe is rotated about line XY to position where reading V_H of Hall probe is maximum.

(a) Hall probe is now moved away from wire, along line XY. Sketch graph to show variation of Hall voltage V_H with distance x of probe from wire

(b) Hall probe is now returned to its original position, distance d from wire. At this point, magnetic flux density due to current in wire is proportional to current.

For direct current of 4.0 A in wire, reading of Hall probe is 3.5 mV. Direct current is now replaced by alternating current of root-mean-square (r.m.s.) value 4.0 A. Period of alternating current is T. Sketch variation with time t of reading of Hall voltage V_H for 2 cycles of alternating current

(c) Student suggests that Hall probe in (a) is replaced with small coil connected in series with a millivoltmeter. Constant current in wire is 4.0 A. In order to obtain data to plot graph showing variation with distance x of magnetic flux density, student suggests that readings of millivoltmeter are taken when coil is held in position at different values of x. Comment on suggestion.

Reference: Past Exam Paper – June 2014 Paper 42 Q5



Solution 461:
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A Hall probe is placed a distance d from a long straight current-carrying wire, as illustrated in Fig. 5.1.