Questions on Potential dividers markscheme

1.       (i)      M marked at the end of the graph                                                                  B1

(ii)     current is 5 (A) and p.d is 6 (V)                                                                     C1

P = VI \ p = 6.0 ´ 5.0

(Allow p = I2 R or p = V2 \ R)                                                                         C1

power = 30 (W)                                                                                             A1

(iii)    1.      VL = 1.0 (V) (From the I/V graph) \ RL = 1.0/2.0 or 0.5 (W)                  M1

VR = 1.2 ´ 2.0 \ RT = 1.2 + 0.5                                                             M1

V = 1.0 + 2.4 \ V = 1.7 ´ 2.0                                                                 A1

voltmeter reading = 3.4 (V)                                                                  A0

2.      Vr = 4.5 - 3.4 (= 1.1 V) \ 4.5 = 2.0r + 3.4 (Possible ecf)                      C1

r = 0.55 (W)             (1.05 W scores 0/2 since the lamp is ignored)       A1

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  2.     (i)      p.d across 1.5 k W resistor = 5.0 - 1.2 = 3.8 (V)                                            B1

(ii)                                   C1

                             C1

R = 474 (W) » 470 (W)                                                                                   A1

(Using 3.8 V instead of 1.2 V gives 4.75 kW - allow 2/3)

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3.       (i)      Any four from:

1. The resistance of the thermistor decreases (as temperature is increased)    B1

2. The total resistance (of circuit) decreases                                                   B1

3. The voltmeter reading increases                                                                 B1

          4. Explanation of 3. above in terms of ‘sharing voltage’
/                                                                             B1

5. The current increases / ammeter reading increases                                     B1

6. Explanation of current increase in terms of                                 B1

          (Allow ecf for statements 3. and 5. if statement 1. is incorrect - maximum
score of 2/4)

(ii)                              C1

                       C1

R = 467 (W) » 470 (W)                                                                                   A1

(When 1.4 V and 3.6 V are interchanged, then R = 3.1 ´ 103 (W) can score2/3)

(Calculation of total circuit resistance of 1.67 ´ 103 (W) can score 2/3)

(Use of  scores 0/3)

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  4.     (i)      (NTC) thermistor                                                                                            B1

(ii)     Resistance decreases when temperature is increased. (ora)                            B1

(Allow correct credit for a PTC thermistor)

(iii)    1       I = (0.032-0.006 =) 0.026 (A)                                                              B1

2       (V200 = 0.026 ´ 200 =) 5.2 (V) / (V720 = 0.006 ´ 700 =) 4.2 (V)           C1

         E = 5.2 – 4.2                  (Allow E = 4.2 – 5.2)                                      C1

         E = 1.0 (V)                    (Allow 1 sf answer)                                        A1

         (9.4 (V) scores 1/3)

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  5.     (i)      The resistance of LDR/circuit changes (as light intensity changes)                 B1
When blade blocks light, resistance of LDR/circuit is large(r) (ora)               B1
Correct statement about p.d (Possible ecf)                                                     B1

(ii)     1.      (V = 5.0 – 3.0)
2.0 (V)        (Allow 1 sf answer)                                                           B1

2.      V =             I = 2.0/2200 / 9.1 ´ 10–4 (A)                           C1
(3.0 = )   (R = 3.0 / 9.1 ´ 10–4)
R = 3300 (W)                 R = 3300 (W)        Possible ecf                        A1
(For VLDR = 2.0 V, R = 1.47 kW. This scores 1/2)
(If 3.5 V given in (ii)1., then R = 940 W. This scores 2/2)

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