LDR in moonlight & thermistor 3

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

[9]

 

  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)

[4]

 

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)

[7]

 

  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)

[6]

 

  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)

[6]

 

 

current in solids answers

Currents in solids

Some copper fuse wire has a diameter of 0.22mm and is designed to carry currents of up to 5.0 A. If there are 1.0 x 10²⁹ electrons per m³ of copper, what is the mean drift speed of the electrons in the fuse wire when it carries a current of 5.0A?

I = nAve rearranging v = I/nAe             Area = ∏r² r = 0.22 x 10^-3 /2 so ∏r² = ∏(0.11 x 10^-3)² = 3.8 x 10^-8 m²

So v = 5.0/ (1.0 x 10²⁹)( 3.8 x 10^-8)(1.6 x 10^-19) = 8.2 x 10^-3 ms^-1

 

A wire carrying an electric current will overheat if there is too large a current: the accepted value for the maximum allowable current in a copper wire is 1.2 x 10⁷ A per square metre of cross-section of the wire. If there are 1.0 x 10²⁹ electrons per m³ of copper, calculate the mean drift speed of the electrons in the wire when the current reaches this value.

I = nAve rearranging v = I/nAe            

So v = 1.2 x 10⁷ / (1.0 x 10²⁹)(1)(1.6 x 10^-19) =  7.5 x 10 ^-4 ms^-1 = 0.75 mms^-1

 

Two copper wires of diameter 2.00 mm and 1.00mm are joined end-to-end. What is the ratio of the average drift speeds of the electrons in the two wires when a steady current passes through them? In which wire are the electrons moving faster?

I = nAve rearranging v = I/nAe             Area = ∏r² r

 

For d = 1.00 mm r = 1 x 10^-3 /2 so ∏r² = ∏(0.5 x 10^-3)² = 7.85 x 10^-7 m²

I don’t know the current so I, but I do know it is the same in both so let’s say it is 1A

So v = 1/ (1.0 x 10²⁹)( 7.85 x 10^-7)(1.6 x 10^-19) = 7.9  x 10^-5ms^-1

For d = 2.00 mm r = 2 x 10^-3 /2 so ∏r² = ∏(1 x 10^-3)² = 3.14 x 10^-6 m²

I don’t know the current so I, but I do know it is the same in both so let’s say it is 1A

So v = 1/ (1.0 x 10²⁹ (3.14 x 10^-6)(1.6 x 10^-19) = 1.99  x 10^-5ms^-1

Ratio of 2mm/1mm = 1.99/7.9 = 0.25 = 1:4

 

Alternate

I = n₁A₁v₁e₁ = n₂A₂v₂e₂           As e and n are constant A₁v₁ = A₂v₂ and v₂/v₁ = A₁/A₂ 

as A = ∏r² = ∏(d/2)²               then A₁ = ∏(1/2)²  and A₂ =∏(2/2)² so A₁/A₂ = ∏(1/2)²/∏(2/2)² = 1/4

 

 

A copper wire joins a car battery to one of the tail lamps and carries a current of 1.8A. The wire has a cross-sectional area of 1.0 mm² (1x 10^-6m²)and is 6.0 m long. If there are 1.0 X 10²⁹ electrons per m³ of copper, calculate how long it takes an electron to travel along this length of wire.

I = nAve rearranging v = I/nAe            

So v = 1.8 / (1.0 x 10²⁹)(1 x 10^-6)(1.6 x 10^-19) = 8.2 x 10^-3 ms^-1 = 1.125 x 10 ^-4 ms^-1

 v = s/t rearranging t = s/v = 6m / 1.125 x 10 ^-4 ms^-1 = 5.33 x 10⁴ s = 14.8 hrs