Question:
Fig. 9.1 shows an electric circuit which includes uninsulated resistance wire XY. A teacher shows some students how to complete the circuit by placing the contact C at various positions on the
wire XY.
(a) The students place contact C at Y. They measure the current on the ammeter. Then they move the contact C along the wire from Y to X.
State and explain the effect on the ammeter reading when they move the contact C from Y to X.
…………………………………………………………………………………………………………………………………
Answer/Explanation
Ans: (current/reading/it) increases
(because circuit) resistance decreases
(b) Calculate the reading on the ammeter when contact C is at X.
ammeter reading = ……………………………………………… A
Answer/Explanation
Ans: 0.75 (A)
6(.0) ÷ 8(.0)
V= IR or (I =) V/R
(c) The students move contact C to point P. The resistance of the wire between X and P is 20 Ω.
Calculate the total resistance of the resistance wire between X and P and the fixed resistor.
total resistance = …………………………………………….. Ω
Answer/Explanation
Ans: 28 (Ω)
(total resistance =) R1 + R2 OR 20 + 8(.0)
d) The electric current in the circuit produces two effects.
Place a tick (3) in the boxes next to these two effects.
Answer/Explanation
Ans: tick in 2nd box (magnetic)
tick in 3rd box (heating)
Question
(a) State the relationship between
(i) the resistance R and the length L of a wire of constant cross-sectional area,
(ii) the resistance R and the cross-sectional area A of a wire of constant length.
(b) A 60 W filament lamp X is connected to a 230 V supply, as shown in Fig. 9.1.
Calculate the current in the filament.
current = …………………………………………..
(c) Lamp Y has a filament made of the same metal as the filament of lamp X in (b).
This filament has half the length and one-third of the cross-sectional area of the filament
of X.
Lamp Y is also connected to a 230 V supply.
Calculate the ratio 
Show your working.
ratio = ……………..
Answer/Explanation
Answer:
(a)(i)(ii) R ∝ L in words or symbols
(ii) AND R ∝ 1/A in words or symbols
(b) P = IV OR (I =) P/V OR 60/230
0.26 A
(c) length change divides resistance by 2/multiplies current by 2
cross-section change multiplies resistance by 3/divides current by 3
(overall) resistance of Y is 3/2 times bigger/3/2 × 885 Ω / 1327 Ω
OR current in Y 2/3 of 0.26A = 0.17A
current in Y/Current in X = 2/3