買四第張五第日三廿月三年展丙摩護
WAH KIÙ, YAT PO
報日僑華
1976 中學會考試題預習專欄
物理科 (廿四)
何辉明
PHYSICS (24)
SOLUTION FOR LAST WEEK
1.Let the resistance of.each
through Dy and Ly are equal
electric bulb be R.
(a)(i) The currents flowing
·and both equal to.
V.
·R
both would give the same brightness..
(ii) The currents flowing.
through I and by are the same and equal tó
頭
both would give the same brightness.
(b)(i) If V.
Vand from the result in (a) the current. flowing through each electric bulb in fig. 1. is 2 times that in: fig..2..
Therefore the electric bulbs.
in fig. 1 is brighter than that in: fig.2.
(ii)Given that V2 =
=2y1
the current through each electric bulb in fig•
V
= 2R
R
the current through each electric bulb in fig. 1.
the electric bulbs in
fig.1 and that in. fig.
would give the same brightness.
(c)(1)The resistance of each
electric bulb =
往台灣
強百貨用品:
(ii)With reference to fig. ∵the\current flowing
through I in fig.1
A
(a)
2. (a)
With reference to fig. 2,
the equivalent resistance of the circuit = 6+6=12:52 the current through In
12
resistor installed
The current through In=
But the current through
the brightness of Ly is not affected but the brightness. of is is reduced.
Rag = 1002.
G
·To convert a galvanometer into an ammeter, a shunt resistance should be connected.
The current through the shunt resistance R =200.0001
= 19.9999 A
The potential different across the shunt resistance
(0.0001)(100) = 0.01 V the shunt resistance required
四期星
日二十月四年六七九一层公年五十六國民輩中育款備造
To convert the galvanometer into a voltmeter, a multiplier.
resistance should be installed.
The potential difference across
the multiplier resistance
= 100 - 0.01 = 99.99 v
The current through the multiplier resistance".
≈ 0.1 mA = 0.0001 A
the multiplier resistance required
99.99- 0.0001
= 999900 N
(c)The current flowing through the galvanometer is 0.06 mA although the current in.the. external circuit is greater than 0.06 mA. The remaining part is conducted away through the shunt resistance. RA ahown in the fig.
G
··06 0.06m
100
008
Original
Scale
(0.08) 80
as
·Scale after
midi
the potential difference between the points to be measured
-80. v
uestion for next week
∴{a} State Lenz's law and Faraday's
law of electromagnetic. Aductione
(1) Determina the polarity of
the magnetic field produced at A when the switch is closed:
(ii) When the switch $ is juat
closed, would a current
be induced in coil Y? If it does, what is the. direction of the induced current in the resistor
Ry?
(iii) With the switch s kept on closed and Ry. being.
unchanged, what changes have been taken place in the: induced current in. COLI Y? (iv) When the resistance of Ry
is increasing, would the" direction and magnitude of the induced current vary?. (v) with the resistance n
unchanged, the coil is. brought nearer to the coil X. What is the direction
of the induced current in: the resistor Ry? (vi) When the velocity of coil
Y towards coll X is increasing, would. the magnitude and direction of the induced current in the coil Y be altered?
(In each answer above, mention:
the law involved in each casé. ) :(a)(i) State the factors that
affects the resistance of a:conducting wire at constant temperature. (ii) What is the relation.
between the resistance of the wire and these factors? State the relation mathe- "matically. What is the
name of the proportional. Pconstant?:
(b) Prove that at constant tem- perature when the length of the conducting wire is... increased by n times by
pulling the resistance of the
conducting wire is increased. by natimes.
END:
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0.01
19.9999
-0.0005 $2
Ansi
(b).
(b).
1000
G
(0·0001)(100)
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