I leave the arithmetic to you! |
This blog contains answers to exercises set for students. While every effort is made to ensure that the information posted is correct, mistakes may occur from time to time.
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Tuesday, December 18, 2018
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Tuesday, November 27, 2018
Resistivity
. (a) either (If in parallel) when one bulb fails, other
bulbs stay on
or (If in parallel) can identify which bulb has failed; (1) 1
or (If in parallel) can identify which bulb has failed; (1) 1
(b) (i) P = VI (1)
0.5 = 240 I
I = 2.1 × 10–3 A 1 s.f. in answer (–1) once only (1) 2
0.5 = 240 I
I = 2.1 × 10–3 A 1 s.f. in answer (–1) once only (1) 2
(ii) R
= V/I (1)
= 240/(2.1 × 10–3)
= 1.14 × 105 Ω or 1.15 × 105 Ω ans
accept (1.1 to 1.2) × 105 Ω. (1) 2
= 240/(2.1 × 10–3)
= 1.14 × 105 Ω or 1.15 × 105 Ω ans
accept (1.1 to 1.2) × 105 Ω. (1) 2
(iii) A
= ρ l / R (1)
= 1.1 × 10–6 × 6.0 × 10–3 / (1.14 × 105) (= 5.79 × 10–14 m2)
A = πr2 (1)
5.79 × 10–14 = πr2 so r = 1.4 × 10–7 m (1) 3
= 1.1 × 10–6 × 6.0 × 10–3 / (1.14 × 105) (= 5.79 × 10–14 m2)
A = πr2 (1)
5.79 × 10–14 = πr2 so r = 1.4 × 10–7 m (1) 3
(iv) filament
too thin / fragile to be manufactured / used without damage;
allow ecf from (iii). (1) 1
allow ecf from (iii). (1) 1
4
[20]
2. (a) current µ p.d / voltage (for a metallic conductor) M1
as
long as temperature is constant / physical conditions remain constant A1
(b) (i) (R
=) (= 0.0349) B1
(ii) R = (Allow
any subject ) C1
C1
resistivity =
5.6 ´ 10–5 A1
unit:
ohm metre / W
m (Allow V m A–1) A1
(5.6 ´ 10–n without unit or incorrect unit and n ¹ 5 or 3 – can
score 2/4)
(5.6
´ 10–3 W m – can score 3/4)
(5.6
´ 10–3 W cm – can score 4/4)
[7]
Monday, November 12, 2018
Tuesday, November 06, 2018
Wednesday, October 31, 2018
Tuesday, October 16, 2018
Ohm's Law calculations
Ohm's Law
I ampere = I A
I milliamp = I mA 1/1000 A
= 10-3A.
I microamp = I μA = 1/1 000 000 A
=10-6 A
I volt = I V
I kilovolt = I kV =1 000 V =103
V
I millivolt = I mV = 1/1000 V = 103 V
1 ohm = 1Ω
1 kilohm = 1KΩ =103 kΩ
1 megohm =
1 MΩ = 1 000 000 Ω = 106Ω
Find the potential difference needed to send a current of 2 μA through a
resistor of 4 kΩ
V=IR V = p.d. in V
I = current in A = 2μA = 2 x 10-6A
R
= resistance in Ω = 4kΩ = 4000Ω
V = 2 x 10-6 x 4000
= 8 X 10-3 V or 8 mV
1 Calculate the current
through a resistor of 20 Ω if the potential difference across it is 0.5 V.
2 When a p.d. of 2 V is
applied across a resistor the current is 0.1 A. Calculate the value of the
resistor?
3 A p.d. of 100 V is applied across a resistor of 2 kΩ. Calculate
the current?
4 Calculate the p.d. that will
produce a current of 2 mA through a resistor of 4 kΩ?
5 Calculate the value of a resistor if a p.d. of 2 V
produces a current of 0.5 μA.
6 Calculate the current through a 1.5 MΩ resistor when a
p.d. of 4.5 V is applied.
7 A lamp operated by a 240 V supply takes a current of
0.625 A. Calculate the resistance of the lamp.
8 A coil of very fine copper wire is found to take a
current of 0.75 A when a p.d. of 4.5 V is applied. If the wire has a resistance
of 1.5 Ω per metre, Calculate length of wire in the coil?
Ohm's Law
I ampere = I A
I milliamp = I mA 1/1000 A
= 10-3A.
I microamp = I μA = 1/1 000 000 A
=10-6 A
I volt = I V
I kilovolt = I kV =1 000 V =103
V
I millivolt = I mV = 1/1000 V = 103 V
1 ohm = 1Ω
1 kilohm = 1KΩ =103 kΩ
1 megohm =
1 MΩ = 1 000 000 Ω = 106Ω
Find the potential difference needed to send a current of 2 μA through a
resistor of 4 kΩ
V=IR V = p.d. in V
I = current in A = 2μA = 2 x 10-6A
R
= resistance in Ω = 4kΩ = 4000Ω
V = 2 x 10-6 x 4000
= 8 X 10-3 V or 8 mV
1 Calculate the current
through a resistor of 20 Ω if the potential difference across it is 0.5 V.
2 When a p.d. of 2 V is
applied across a resistor the current is 0.1 A. Calculate the value of the
resistor?
3 A p.d. of 100 V is applied across a resistor of 2 kΩ.
Calculate the current?
4 Calculate the p.d. that
will produce a current of 2 mA through a resistor of 4 kΩ?
5 Calculate the value of a resistor if a p.d. of 2 V
produces a current of 0.5 μA.
6 Calculate the current through a 1.5 MΩ resistor when a
p.d. of 4.5 V is applied.
7 A lamp operated by a 240 V supply takes a current of
0.625 A. Calculate the resistance of the lamp.
8 A coil of very fine copper wire is found to take a
current of 0.75 A when a p.d. of 4.5 V is applied. If the wire has a resistance
of 1.5 Ω per metre, Calculate length of wire in the coil?
Tuesday, September 25, 2018
Tuesday, September 18, 2018
Monday, September 17, 2018
Wednesday, September 12, 2018
Tuesday, September 11, 2018
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