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.}