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Mark schemes
A
1 [1]
A
2 [1]
D
3 [1]
D
4 [1]
D
5 [1]
(a) the (total) energy transferred/work done when
6 one unit/coulomb of charge
B1
is moved around a circuit/provided by the supply
B1
2
(b) work is done inside the battery/there is resistance
inside the battery
B1
so less energy is available for the external circuit/someoltage
is lost between the terminal/mention of lost volts
B1
2
(c) (i) 9.00 V
c.a.o.
B1
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(ii) lost voltage = E - V or E = I(R + r)
C1
0.82r = 0.59
C1
5
internal resistance = 0.720 Ω
A1
(iii) because the battery has to provide more energy/power
B1
[9]
(a) V = –Ir + (1)
7 1
(b) straight line (within 1st quadrant) (1)
negative gradient (1)
2
(c) : intercept on voltage axis (1)
r: gradient (1)
2
[5]
(a) work done per unit charge
8
Allow V=W/Q if W and Q defined
1
(b) Voltmeter reading / terminal pd drops
Battery has internal resistance
pd occurs within battery / ’lost volts’ within battery / emf is shared between internal and
external resistances
3
[4]
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(a) 12 * 15/25
9
C1
= 7.2 V
A1
2
(b) total R now 32.5
C1
12 * 7.5/32.5 = 2.7[7] V or calculates I = 0.369 A
C1
terminal p.d. 12 ‑ 2.8 = 9.2 V or V = 0.369 × (10 + 15) = 9.2 V
A1
3
[5]
(a) (i) work done (by the battery) per unit charge (1)
10 or (electrical) energy per unit charge
or pd/voltage when open circuit/no current
(ii) the resistance of the materials within the battery (1)
or hindrance to flow of charge in battery
or loss of pd/voltage per unit current
2
(b) (i) (use of E = V + Ir)
12 = V + 800 × 0.005 (1) (working/equation needs to be shown)
V = 12 – 4 = 8.0V (1)
(ii) (use of P = I2r)
P = 8002 × 0.005 (1) (working/equation needs to be shown)
P = 3200 (1) W (1) or J s–1
5
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Mark schemes
A
1 [1]
A
2 [1]
D
3 [1]
D
4 [1]
D
5 [1]
(a) the (total) energy transferred/work done when
6 one unit/coulomb of charge
B1
is moved around a circuit/provided by the supply
B1
2
(b) work is done inside the battery/there is resistance
inside the battery
B1
so less energy is available for the external circuit/someoltage
is lost between the terminal/mention of lost volts
B1
2
(c) (i) 9.00 V
c.a.o.
B1
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, theonlinephysicstutor.com
(ii) lost voltage = E - V or E = I(R + r)
C1
0.82r = 0.59
C1
5
internal resistance = 0.720 Ω
A1
(iii) because the battery has to provide more energy/power
B1
[9]
(a) V = –Ir + (1)
7 1
(b) straight line (within 1st quadrant) (1)
negative gradient (1)
2
(c) : intercept on voltage axis (1)
r: gradient (1)
2
[5]
(a) work done per unit charge
8
Allow V=W/Q if W and Q defined
1
(b) Voltmeter reading / terminal pd drops
Battery has internal resistance
pd occurs within battery / ’lost volts’ within battery / emf is shared between internal and
external resistances
3
[4]
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, theonlinephysicstutor.com
(a) 12 * 15/25
9
C1
= 7.2 V
A1
2
(b) total R now 32.5
C1
12 * 7.5/32.5 = 2.7[7] V or calculates I = 0.369 A
C1
terminal p.d. 12 ‑ 2.8 = 9.2 V or V = 0.369 × (10 + 15) = 9.2 V
A1
3
[5]
(a) (i) work done (by the battery) per unit charge (1)
10 or (electrical) energy per unit charge
or pd/voltage when open circuit/no current
(ii) the resistance of the materials within the battery (1)
or hindrance to flow of charge in battery
or loss of pd/voltage per unit current
2
(b) (i) (use of E = V + Ir)
12 = V + 800 × 0.005 (1) (working/equation needs to be shown)
V = 12 – 4 = 8.0V (1)
(ii) (use of P = I2r)
P = 8002 × 0.005 (1) (working/equation needs to be shown)
P = 3200 (1) W (1) or J s–1
5
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Page 39 of 57
