Ch 13 Resistance Worksheet Answers 353tn
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13 Marking scheme: Worksheet 1 Due to the current flowing through the chemicals within the cell.
[1]
2 e.m.f. of power supply = terminal p.d. + p.d. across internal resistance
[1]
3 a
E = I (R + r) R = 0 (since supply is shorted-out)
so I =
E r
12 = 5.2 A 2.3 b Excessive heating of the power supply.
I=
4 a
[1] [1] [1]
p.d. across internal resistance = E − V = 1.5 − 0.85
[1]
p.d. across internal resistance = 0.65 V
[1]
b I= r= 5 Vout =
V 0.85 = = 0.327 A R 2.6
[1]
0.65 ≈ 2 .0 Ω 0.327
[1]
R2 × Vin R1 + R2
[1]
6 .0 × 5 .0 18 + 6.0
[1]
V = 1.25 V ≈ 1.3 V
[1]
V=
6 a
p.d. across internal resistance = E − V = 1.4 − 0.81
[1]
p.d. across internal resistance = 0.59 V
[1]
b I= r= c
V 0.81 = = 0.111 A R 7.3
[1]
0.59 = 5.3 Ω 0.111
[1]
P = I2R
[1] 2
0.111 × 7.3 0.1112 × (7.3 + 5.3) ratio = 0.58 (42% of the power is ‘lost’ internally in the cell)
Ratio =
7 a
E R+r total e.m.f. = 2 × 1.4 = 2.8 V and total internal resistance = 2 × 0.38 = 0.76 Ω 2 .8 I= = 1.094 A ≈ 1.1 A 0.76 + 1.8
I=
V = IR = 1.094 × 1.8 = 1.97 V ≈ 2.0 V
[1] [1] [1] [1] [1]
b V = E − Ir
[1]
V = 1.4 − (1.094 × 0.38) = 0.98 V
COAS Physics 1 Teacher Resources
[1]
[1]
Original material © Cambridge University Press 2005, 2008
1
13 Marking scheme: Worksheet
8 a
R2 × Vin R1 + R2
[1]
200 × 6.0 180 × 200
[1]
Vout =
V=
V = 3.16 V ≈ 3.2 V b As the resistance decreases, the p.d. across the variable resistor decreases.
[1] [1]
Correct values marked Correct curve shape
[1] [1]
9 Maximum p.d. across the 3.6 kΩ resistor is when the resistance of the thermistor is minimum. [1] R2 [1] Vout = × Vin R1 + R2 3600 × 5.0 430 + 3600 V = 4.47 V ≈ 4.5 V
Vout =
10 a
R/Ω
I/A
[1] [1] V/V
P/W
0.00
3.00
0.00
0.00
0.10
2.50
0.25
0.63
0.20
2.14
0.43
0.92
0.30
1.88
0.56
1.05
0.40
1.67
0.67
1.11
0.50
1.50
0.75
1.13
0.60
1.36
0.82
1.12
0.70
1.25
0.88
1.09
0.80
1.15
0.92
1.07
0.90
1.07
0.96
1.03
1.00
1.00
1.00
1.00
All columns correctly calculated. b The completed table will show that maximum power is dissipated when the external resistor has resistance of 0.50 Ω. This resistance value is equal to the internal resistance. If you want to deliver maximum power to an external load, then its resistance must ‘match’ the resistance of the supply (the cell in this case). Suitable sketch graph.
COAS Physics 1 Teacher Resources
Original material © Cambridge University Press 2005, 2008
[2] [1] [1] [1]
2
[1]
2 e.m.f. of power supply = terminal p.d. + p.d. across internal resistance
[1]
3 a
E = I (R + r) R = 0 (since supply is shorted-out)
so I =
E r
12 = 5.2 A 2.3 b Excessive heating of the power supply.
I=
4 a
[1] [1] [1]
p.d. across internal resistance = E − V = 1.5 − 0.85
[1]
p.d. across internal resistance = 0.65 V
[1]
b I= r= 5 Vout =
V 0.85 = = 0.327 A R 2.6
[1]
0.65 ≈ 2 .0 Ω 0.327
[1]
R2 × Vin R1 + R2
[1]
6 .0 × 5 .0 18 + 6.0
[1]
V = 1.25 V ≈ 1.3 V
[1]
V=
6 a
p.d. across internal resistance = E − V = 1.4 − 0.81
[1]
p.d. across internal resistance = 0.59 V
[1]
b I= r= c
V 0.81 = = 0.111 A R 7.3
[1]
0.59 = 5.3 Ω 0.111
[1]
P = I2R
[1] 2
0.111 × 7.3 0.1112 × (7.3 + 5.3) ratio = 0.58 (42% of the power is ‘lost’ internally in the cell)
Ratio =
7 a
E R+r total e.m.f. = 2 × 1.4 = 2.8 V and total internal resistance = 2 × 0.38 = 0.76 Ω 2 .8 I= = 1.094 A ≈ 1.1 A 0.76 + 1.8
I=
V = IR = 1.094 × 1.8 = 1.97 V ≈ 2.0 V
[1] [1] [1] [1] [1]
b V = E − Ir
[1]
V = 1.4 − (1.094 × 0.38) = 0.98 V
COAS Physics 1 Teacher Resources
[1]
[1]
Original material © Cambridge University Press 2005, 2008
1
13 Marking scheme: Worksheet
8 a
R2 × Vin R1 + R2
[1]
200 × 6.0 180 × 200
[1]
Vout =
V=
V = 3.16 V ≈ 3.2 V b As the resistance decreases, the p.d. across the variable resistor decreases.
[1] [1]
Correct values marked Correct curve shape
[1] [1]
9 Maximum p.d. across the 3.6 kΩ resistor is when the resistance of the thermistor is minimum. [1] R2 [1] Vout = × Vin R1 + R2 3600 × 5.0 430 + 3600 V = 4.47 V ≈ 4.5 V
Vout =
10 a
R/Ω
I/A
[1] [1] V/V
P/W
0.00
3.00
0.00
0.00
0.10
2.50
0.25
0.63
0.20
2.14
0.43
0.92
0.30
1.88
0.56
1.05
0.40
1.67
0.67
1.11
0.50
1.50
0.75
1.13
0.60
1.36
0.82
1.12
0.70
1.25
0.88
1.09
0.80
1.15
0.92
1.07
0.90
1.07
0.96
1.03
1.00
1.00
1.00
1.00
All columns correctly calculated. b The completed table will show that maximum power is dissipated when the external resistor has resistance of 0.50 Ω. This resistance value is equal to the internal resistance. If you want to deliver maximum power to an external load, then its resistance must ‘match’ the resistance of the supply (the cell in this case). Suitable sketch graph.
COAS Physics 1 Teacher Resources
Original material © Cambridge University Press 2005, 2008
[2] [1] [1] [1]
2
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