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/Q38

A 20 V d.c. supply is connected to a circuit consisting of five resistors L, M, N, P and Q.

There is a potential drop of 7 V across L and a further 4 V potential drop across N.

What are the potential drops across M, P and Q?

Solution:

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.

Reference: PYQ - Oct/Nov 2013 Paper 13 Q38

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/Oct Nov/13/2017/Q37

Three identical cells each have electromotive force (e.m.f.) E and negligible internal resistance.
The cells are connected to three identical resistors, each of resistance R, as shown.

What is the potential difference between P and Q?

Solution:

Answer: C

Two of the cells are providing an e.m.f. in one direction and the other in the opposite direction. So,

Overall e.m.f. = E + E – E = E

The resistors are connected in series.

Total resistance = R + R + R = 3R

From Ohm’s law, I = V / R

Current in circuit = overall e.m.f. / total resistance = E / 3R

Consider the potentials in the section PQ.

Let P be at a potential of E.

The e.m.f. of the cell is E. That is, the potential difference between the terminals of the cell is E. If point P is at a potential of 0, then the point between the positive terminal of the battery and the resistor is E.

Current in the circuit = E / 3R

p.d. V across a resistor = IR = (E/3R) × R = E / 3

Thus, potential at point Q = E – E/3 = 2E / 3

In another wording, potential difference between 2 points means the ‘difference in potential’ between the 2 points.

p.d. between P and Q = potential at Q – potential at P

p.d. between P and Q = (2E / 3) – 0 = 2E / 3

Reference: PYQ - Oct/Nov 2017 Paper 13 Q37

9702/Oct Nov/13/2017/Q36

Two cells, each with electromotive force (e.m.f.) E, but different internal resistances r1 and r2, are
connected in series to a resistor R. The reading on the voltmeter is 0 V.

What is the resistance of R?

Solution:
Answer: B


Total e.m.f = E+E = 2E

Total resistance = r1 + r2 + R

Current, i = (2E)/ (r1+r2+R) ------------------(i)

P.D. across 1st cell

V = E - ir1

0 = E - ir1

This implies,

i = E/r1 ------------------(ii)

From (i) and (ii), we have

=> r1 + r2 + R = 2r1

R = r1 - r2

Reference: PYQ - Oct/Nov 2017 Paper 13 Q36

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/Oct Nov/11/2011/Q33

Which statement about electrical resistivity is correct?

A The resistivity of a material is numerically equal to the resistance in ohms of a cube of that

material, the cube being of side length one metre and the resistance being measured

between opposite faces.

B The resistivity of a material is numerically equal to the resistance in ohms of a one metre

length of wire of that material, the area of cross-section of the wire being one square

millimetre and the resistance being measured between the ends of the wire.

C The resistivity of a material is proportional to the cross-sectional area of the sample of the

material used in the measurement.

D The resistivity of a material is proportional to the length of the sample of the material used in

the measurement.

Solution:
Answer: A

Let's take this option by option, with the following formula in focus:

ρ = RA/l

Where ρ = resistivity of the material through which current is being passed,
R = resistance of the sample through which current is being passed,
A = Area of the cross section of the sample through which current is being passed (perpendicular to the direction of current), and
l = length of cross section of sample through which current flows (i.e. the distance through which the current flows in the sample used).

So, for option A, let's see what the formula tells us:

If the cube has a one meter side and the current is passed from one face to another, we can say that the current travels 1 meter from one end of the cube to the other; therefore, l = 1 meter. Alternatively, the current first enters the cube on one side, travels 1 meter, and exits the cube, giving us the same value.

Further, the cross section through which the current travels, perpendicular to the direction of current - suppose you place the cube on the table, and cut it parallel to the edges, the area you get after the cut is an area of 1 meter x 1 meter = 1 m^2.
Another way of getting this is to imagine there is no current passing through the cube. You close the circuit, and current starts flowing. Soon, it gets to the beginning of the cube, and starts moving through it. Suppose no charge carrier (electron/proton, either is fine) travels faster than another, you have a "wall" of such charge carriers advancing along the cube. Now ask yourself. How large is that wall? What is the area of that wall? In this case, the area of that wall is 1 m^2, which is our result.

Putting these in the equation, we get

ρ = R * 1 m^2/1 m = R
So ρ = R

Therefore, A is right - the value of ρ is numerically equal to the value of the resistance of the cube in this situation.

Option B: Doing the maths here again, we can say that the length through which the current passes is l = 1 meter and the cross sectional area through which the current passes is 1 mm^2 = (1/1000 meters)^2 = (10^-3)^2 = 10^-6.
Putting it in the equation,

ρ = R * 10^-6/1 = 10^-6 * R
Which is not right.

Option C: the option says that resistivity is dependent on the cross sectional area of the sample used. Even though the formula says that ρ = R/l * A, the wording of the option is something very precise; when it says proportional, what it means is

"...a change in A results in a change in ρ, such that the change in A can be equated to the corresponding change in ρ if a suitable constant k is introduced as a factor of change (i.e. ΔA = k * Δρ)"

This is, of course, false. At a fixed temperature, given no other external electric or magnetic interference (literally and figuratively), the resistivity of a material will not change with any change in area. Instead, the resistance will change to ensure that the value of ρ remains the same. Since resistivity does not depend on the dimensions of a sample and only on the innate nature of the material, C cannot be right.

Changing the cross-sectional area of a sample may change the resistance, but it cannot change the resistivity.

Option D: The same argument as option C can be supplied here, replacing Cross Sectional Area A with length l.

Reference: PYQ - May/Jun 2011 Paper 11 Q33

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

9702/May Jun/12/2015/Q3

An analogue ammeter has a pointer which moves over a scale. Following prolonged use, the
pointer does not return fully to zero when the current is turned off and the meter has become less sensitive at higher currents than it is at lower currents.

Which diagram best represents the calibration graph needed to obtain an accurate current
reading?

Solution:

Answer: C.

The scale reading will not start from "0" since the pointer does not return fully to zero when the current is turned off. So choice "A" and "B" are not the answer.

For choice D, the curve line of the graph shows the higher current get higher scale reading. Which means the higher current is more sensitive than lower current.

Only choice C shows the correct reading and fulfill the conditions of the question mentioned.

Reference: PYQ - May/Jun 2015 Paper 12 Q3

9702/May Jun/11/2015/Q38

A wire RST is connected to another wire XY as shown.

Each wire is 100 cm long with a resistance per unit length of 10 Ω m^–1.
What is the total resistance between X and Y?
A 3.3 Ω
B 5.0 Ω
C 8.3 Ω
D 13.3 Ω

Solution:

Answer: C.

Resistance per unit length of wire = 10 Ω m^–1

Resistance of wire RST = 1m × 10 Ω m^–1 = 10 Ω

Since the wire XY is 100cm, the sum of XR + TY = 100 – 50 = 50cm

Sum of resistance of (XR+TY) = 0.5m × 10 Ω m^–1 = 5 Ω

Resistance of wire RT = 0.5m × 10 Ω m^–1 = 5 Ω

Hence the arrangement is effectively a 10 Ω resistor RST in parallel with a 5 Ω resistor RT, and in series with another 5 Ω resistor (XR+TY).

Effective resistance of parallel combination = [1/10 + 1/5]^-1 = 3.3 Ω

The total resistance is therefore 5 Ω + 3.3 Ω = 8.3 Ω.

Reference: PYQ - May/Jun 2015 Paper 11 Q38