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Question 762: [Current of Electricity]

In the circuit below, ammeter reading is I and voltmeter reading is V.

When switch is closed, which row describes what happens to I and V?

I                                   V

A         decreases                     decreases to zero

B         increases                      decreases to zero

C         increases                      stays the same

D         stays the same             increases

Reference: Past Exam Paper – June 2012 Paper 11 Q36 & Paper 13 Q35

Solution 762:

Answer: B.

When the switch is opened, current passes through both resistors in the circuit.

Let the resistance of one resistor be R. Assume they are of the same resistance.

When switch is opened, Total resistance R_T in circuit = R + R = 2R

Ohm’s law: V = IR

This gives a current I (= E / R_T where E is the e.m.f. of the supply). The total e.m.f. in the circuit is divided as p.d. across each of the resistor.

When the switched is closed, the resistor connected in parallel to the wire is short circuited. That is, all current would pass through the wire (of negligible resistance) instead of passing through that resistor.

When switch is closed, Total resistance R_T in circuit = R + 0 = R

This gives a current I (= E / R_T where E is the e.m.f. of the supply) which is now greater than before since R_T is less.

Ohm’s law for the resistor: V = IR

Since the current I through the resistor (in parallel) is zero, the p.d. across it is also zero.

Question 763: [Current of Electricity]

A cell, two resistors of equal resistance and ammeter are used to construct four circuits. The resistors are the only parts of the circuits that have resistance.

In which circuit will ammeter show the greatest reading?

Reference: Past Exam Paper – November 2011 Paper 11 Q37 & Paper 13 Q36

Solution 763:

Answer: D.

Ohms’ law: V = IR

Current I in the whole circuit = V / R_T where R_T is the combined resistance of the resistors.

Let the resistance of 1 resistor = R. Let the e.m.f. of the supply = E

Circuit A consists of a series connection of the 2 resistors. So, R_T = R + R = 2R. Current I = E / 2R. Here, the current will be the lowest.

To measure the current flowing through a resistor, the ammeter, which has a negligible resistance, must be connected in series with it. If it is connected in parallel with a component, the current would flows through it instead of the component (since the ammeter has zero resistance) and that component would be short circuited.

In circuit B, R_T = R + 0 = R. Current I = E / R.

Circuit C and D both consists of a parallel combination of the 2 resistors.

Combined resistance R_T in circuit = [1/R + 1/R]^-1 = R / 2

Current I in circuit = E / (R/2) = 2 (E / R)

These 2 circuit contain the greatest current of all the 4 choices. However, at a junction, the current splits. So, the ammeter in circuit C gives only half the total current in the circuit.

Current flows from the positive terminal of the supply to its negative terminal. In choice D, the ammeter is connected directly to the positive terminal, so it reads the value of the current flowing in the whole circuit.

Question 764: [Dynamics > Collisions]

Diagram shows a particle X, with kinetic energy E_k, about to collide with a stationary particle Y.

Both particles have the same mass.

After colliding, X and Y travel onwards together as a single larger particle.

How much kinetic energy is lost in collision?

A 0                  B E_k / 4                        C E_k / 2                        D 3E_k / 4

Reference: Past Exam Paper – November 2014 Paper 13 Q16

Solution 764:

Answer: C.

In any closed system, momentum is always conserved.

Before collision,

Let speed of X be v. Let mass of X = mass of Y = m

Momentum of X = mu

Momentum of Y = m(0) = 0

After collision,

Both particles travel as a single larger particle.

Mass = m + m = 2m

Let the velocity after collision be v.

Total momentum = (2m)v = 2mv

From the conservation of momentum,

2mv = mu

Velocity v = u / 2

Kinetic energy before collision, E_k = ½ mu²

Kinetic energy after collision = ½ (2m) (u/2)² = ¼ mu²

Lost in Kinetic energy = ½ mu² – (¼ mu²) = ¼ mu² = E_k / 2

Question 765: [Current of Electricity > Potential divider]

In circuit shown, XY is a length L of uniform resistance wire. R₁ and R₂ are unknown resistors. J is a sliding contact that joins the junction of R₁ and R₂ to points on XY through small signal lamp S.

To determine ratio V₁ / V₂ of the potential differences across R₁ and R₂, a point is found on XY at which the lamp is off. This point is at a distance x from X.

What is value of the ratio V₁ / V₂?

A L / x                         B x / L                         C (L – x) / x                            D x / (L – x)

Reference: Past Exam Paper – June 2011 Paper 11 Q37 & Paper 13 Q36

Solution 765:

Answer: D.

For the lamp to be off, the potential difference across it should be zero so that no current flows through it. That is, the potential at the junction of R₁ and R₂ should be equal to the potential at the sliding contact J

The p.d. across R₁ is V₁ and the p.d. across R₂ is V₂. So, V₁ should be equal to the p.d. across XJ and V₂ should be equal to the p.d. across JY.

Since point X is connected to the negative terminal of the supply, it can be considered to be at a potential of 0. Point Y is at the maximum potential, equal to the e.m.f. of the supply.

The wire XY, together with the sliding contact J, can be considered as a potential divider.

Resistance R of the wire = ρL / A

Since ρ and A are assumed to be constant, the resistance of the wire is proportional to the length L.

Ohm’s law: V = IR

The p.d. across the wire is proportional to R, which is itself proportional to L. So, the p.d. across the wire is proportional to the length of wire being considered.

A length of L corresponds to a p.d. equal to the e.m.f. E of the supply

A length of x corresponds to a p.d. equal to (E / L) × x = (x / L) × E

Note that this is a p.d. (potential difference – or in other words, a difference in potential).  Since X is at a potential of zero, point J would be at a potential of [(x / L) × E] – 0 = (x / L) × E.

V₁ = (x / L) × E

Now, the part JY of the wire corresponds to a length (L – x) of the wire.

A length of (L – x) corresponds to a p.d. equal to (E / L) × (L – x) = [(L – x) / L] × E

V₂ = [(L – x) / L] × E

Ratio = V₁ / V₂ = x / (L – x)

Question 766: [Pressure]

Atmospheric pressure at sea level has value of 100 kPa.

Density of sea water is 1020 kg m^–3.

At what depth in the sea would total pressure be 110 kPa?

A 1.0 m                       B 9.8 m                       C 10 m                        D 11 m

Reference: Past Exam Paper – June 2010 Paper 11 Q17 & Paper 12 Q18 & Paper 13 Q15

Solution 766:

Answer: A.

The pressure at sea level is 100 kPa due to the atmosphere (atmospheric pressure).

The pressure only due to a liquid column of depth h and density ρ is hρg.

Total pressure at a depth h = Atmospheric pressure + hρg

110 000 = 100 000 + (h × 1020 × 9.81)

Depth h = 10 000 / (1020 × 9.81) = 0.999 = 1.0m