Wires are used to connect a battery of negligible internal resistance to a lamp, as shown in Fig. 7.1.

Question 19

(a) Define the ohm. [1]

 

 

(b) Wires are used to connect a battery of negligible internal resistance to a lamp, as shown in Fig. 7.1.

Fig. 7.1

 

The lamp is at its normal operating temperature. Some data for the filament wire of the lamp and for the connecting wires of the circuit are shown in Fig. 7.2.

 

Fig. 7.2

 

(i) Show that

resistance of filament wire     

total resistance of connecting wires = 1000.

[2]

 

 

(ii) Use the information in (i) to explain qualitatively why the power dissipated in the filament wire of the lamp is greater than the total power dissipated in the connecting wires. [1]

 

 

(iii) The lamp is rated as 12 V, 6.0 W. Use the information in (i) to determine the total resistance of the connecting wires. [3]

 

 

(iv) The diameter of the connecting wires is decreased. The total length of the connecting wires and the resistivity of the metal of the connecting wires remain the same.

 

State and explain the change, if any, that occurs to the resistance of the filament wire of the lamp. [3]

[Total: 10]

 

 

 

 

 

Reference: Past Exam Paper – November 2017 Paper 21 Q7

 

 

 

 

 

Solution:

(a) (the ohm is defined as) volt / ampere

 

 

(b)

(i)

{Resistance of wire:}

R = ρL / A

 

{Cross-sectional area A = πd² / 4

 

Resistance of filament wire = ρL / (πd² /4)

Resistance of connecting wires = 0.028ρ × 7.0L / {π (14d)² / 4}

}

 

ratio = [ρL / (πd² /4)] / [0.028ρ × 7.0L / {π (14d)² / 4}] = 1000

 

or

ratio = 14² / (0.028 × 7) = 1000

 

 

(ii)

The same current flows (in the connecting and filament wires) but the lamp/filament (wire) has greater resistance

 

{Power dissipated = I²R}

 

 

(iii)

P = V² / R                    or P = VI                      or P = I²R

 

{Make R the subject formula in whichever you choose.

 

R = V² / P                    or P = I² / R                 or R = V / I  and I = P / V = 6/12 = 0.5 A}

 

(for filament wire) R = 12² / / 6.0        or R = 6.0 / 0.50²        or R = 12 / 0.50

(for filament wire) R = 24 Ω

 

{From the ratio of (b)(i), the ratio of

Resistance of filament wire / total resistance of connecting wires = 1000

 

Total resistance of connecting wires = resistance of filament wire / 1000

Total resistance of connecting wires = 24 / 1000}

 

(for connecting wire) R = 24 / 1000 = 2.4 × 10^–2 Ω

 

 

(iv)

{Resistance of wire: R = ρL / A

When the diameter is decreased, the cross-sectional area A decreases.}

 

The resistance of the connecting wire increases.

 

{The total resistance in the circuit increases.

From Ohm’s law, I = V / R

When the resistance increases, the current in the circuit decreases.}

 

The current in circuit/lamp/filament (wire) decreases.

or

potential difference across lamp/filament (wire) decreases

 

{Now, considering the FILAMENT lamp.

V = IR

Since the current has decreased, the p.d. across the lamp decreases.}

 

(So) the resistance of lamp/filament (wire) decreases.

 

{When the p.d. across a filament lamp increases, its resistance also increases. In the same way, when the p.d. across the filament lamp decreases, its resistance also decreases.}

 

A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig. 2.1.

Question 10

(a) (i) State the basic assumption of the kinetic theory of gases that leads to the conclusion that the potential energy between the atoms of an ideal gas is zero. [1]

 

(ii) State what is meant by the internal energy of a substance. [2]

 

(iii) Explain why an increase in internal energy of an ideal gas is directly related to a

rise in temperature of the gas. [2]

 

 

(b) A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig. 2.1.

 

Fig. 2.1

(i) State the change in internal energy of the gas during one complete cycle PQRP. [1]

 

(ii) Calculate the work done on the gas during the change from P to Q. [2]

 

(iii) Some energy changes during the cycle PQRP are shown in Fig. 2.2.

 

Fig. 2.2

Complete Fig. 2.2 to show all of the energy changes. [3]

 

 

Reference: Past Exam Paper – November 2010 Paper 41 & 42 Q2

 

Solution:

(a) (i) There are no forces (of attraction or repulsion) between the atoms / molecules / particles.

 

(ii) The internal energy of a substance is the sum of kinetic and potential energy of the atoms / molecules due to their random motion.

 

(iii) The (random) kinetic energy increases with temperature. There is no potential energy. (since the gas is ideal)

So, an increase in temperature increases the internal energy.

 

 

(b)

(i) Change in internal energy = Zero

{The initial and final states (values of p and V) are identical in a complete cycle, so the change in internal energy is zero.}

 

(ii) Work done = pΔV = 4.0×10⁵ × 6×10^-4 = 240 J

 

(iii)

(values for the change in internal energy should add up to zero.)

{ΔU = ΔQ + ΔW

+ΔQ is the amount of heat/energy supplied to the gas.

+ΔU is the increase in internal energy.

+ ΔW is the work done ON the gas.   ΔW = p ΔV

For P to Q: ΔQ = - 600 J. As calculate above, ΔW = + 240 J. ΔU = - 600 + 240 = - 360 J

For Q to R: ΔQ = +720 J. ΔV = 0, so ΔW = 0. ΔQ = ΔU – 0 = +720 J

 

For R to P: ΔQ = +480 J.

As answered in (b)(i), the change in internal energy during one complete cycle is zero.

Let the change in internal energy for this change be ΔU. Consider the changes in internal energy for P to Q and Q to R.

-360 + 720 + ΔU = 0.

So, ΔU = -360 J.

ΔW = ΔU – ΔQ = -360 – 480 = -840 J

Note that from R to P, both the pressure and the volume are changing. So, we cannot use the simple equation of W = pΔV, which assumes a constant pressure.}