2009 direct current circuits tutorial revised solutions

7/30/2019 2009 Direct Current Circuits Tutorial Revised Solutions

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Tutorial 14 DC Circuit Tutorial Suggested Solutions

1. a) i) I1 = I4 (Principle 1)

ii) I2 = I1 + I3 (Principle 1)

iii) E1 = V1 + V2 (Principle 2)

iv) E2 = V2 + V3 (Principle 2)

v) E1 - V1 = E2 - V3 (Principle 2) [Alternatively, use (iii) and (iv) to derive (v)]

2. a) The battery is flat.

b) Lamp As filament has blown.

c) There is a broken connection between the battery and lamp A or between the batteryand lamp F.

3.

4. 12 I(3.0) = 0.5 (I) (6.0)

I= 2 A whereI is the total current

Hence the ammeter reading is 1 A.

Note: The total current supplied by the battery will be split equally at the junction.

[Alternatively, try simplifying the circuit using series and parallel connection]

5. When variable resistor is 0 k,

Rc = 1.0 k.

Voltmeter reading = 12 V (maximum)

When variable resistor is 1.0 k,

Voltmeter reading = 1)11(1

1++

12 = 8 V (minimum)

6. Rc =

1

31

21

11

++ = 0.545

2

-5

-8

0

V / V

P Q R S


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7. RPQ =

1

9

1

1

1

+ = 0.9 ; RPR =

1

7

1

3

1

+ = 2.1 ; RPS =

1

4

1

6

1

+ = 2.4

RQR =

1

81

21

+= 1.6 ; RQS =

1

51

51

+= 2.5 ; RRS =

1

31

71

+= 2.1

Resistance is maximum between Q and S.

8. Answer: D

9. Answer: B

10. Total R = R +

1

R

1

R

1

+= R + 0.5 R = 1.5 R

V =R5.1

0.5R 6 = 2 V

Hence the voltmeter reading is 2 V.

11. Rc = 20 + 50 +

1

12000

1

800

1

+ = 820

Ammeter reading, I =820

6= 7.32 mA

Voltmeter reading = 6 (0.00732 20) - (0.00732 50) = 5.49 V

[Alternatively, try potential divider method and V = IR method]

12. Vat Y =LDR

RR

R

+ 6

2 =1200+RR

6

R = 600

13. a) i) R =

1

3000

1

600

1

+ = 500

ii) R =

1

600

1

600

1

+ = 300

iii) R =

11

600

1

+ = 600


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b)

c) i) 1) Reffof the 600 resistor and LDR = (3000

1

600

1+ )-1 = 500

Current supplied by battery =50030

12

+= 0.02264 A

Voltage across LDR = 12 (0.02264 30) = 11.32 V

Therefore,

Current through LDR =3000

32.11= 3.77 mA

2) Power dissipated in LDR = 0.00377 11.32 = 0.0427 W (Try P = I2R too)

c) ii) When RLDR drops to 100 ,

Current supplied by battery =1

100

1

600

1

30

12

++= 0.1037 A

Voltage across LDR = 12 (0.1037 30) = 8.889 V

ILDR = 08889.0100

889.8= A

Power dissipated in LDR = 0.088892 100 = 0.79 W > 0.5 W [Try P = VI too]

Hence, LDR will be overheated and damaged.

d)

600

3000

300

500

600

0

R /

R2

/

30

600

12.0 V

E


4/8

14. a) R =A

=

( )( )2

6

00055.0

2.1101.1

= 1.389 = 1.4

b) P.d. across XY = 7.04.1

4.1

+ 3.0 = 2 V

p.d. per unit length of XY =2.1

2= 1.666 = 1.7 V m-1

c) i)

ii) Position of J on XY =7.1

5.1= 0.9 m away from X

iii) Change the material of the wire to one with a lower resistivity OR larger cross

sectional area.

15. a) i) The fixed resistor R forms a potential divider with the thermistor. Without the fixedresistor, the voltmeter will always read 9.0 V, regardless of the temperature of thethermistor (assume voltmeter is put across thermistor).

b) i) 1.2 k

ii) V = 0.92.10.5

0.5

+= 7.258 = 7.3 V

c) Between 2.5 oC and 30 oC, the voltmeter reading only varies from 5.3 V to 7.3 V,

which is a narrow range. Hence, the voltmeter is not able to show small temperaturechanges.

16. a) The thermal energy in a body depends on the sum of the k.e. and p.e. of all themolecules of the body (the unit is joule). However, temperature is only a measure ofhow hot or cold an object is (the unit for temperature is Kelvin). Hence temperature isnot a measure of the quantity of thermal energy in a body.

b) i) The variation of the resistance of the thermistor is much greater and morelinear for temperature between 273 K to 293 than for temperature between 313 K to333 K. Hence, the thermometer is more sensitive in the range 273 K to 293 K.

ii) T = 292 K (approx.)

Current through cell C


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c) i) VA = 6400600

600

+= 3.6 V

For voltmeter to read 0 V, VB has to be at 3.6 V as well. Hence,

VB = 61200

1200 +T

R

3.6 = 61200

1200

+T

R

RT = 800

Therefore, T = 305 K (from the graph)

ii) 1) When voltmeter read 1.2 V, it suggests that the voltage at point B could be either 1.2V higher or lower than the voltage at A. Hence, the thermistor could be at one of

these 2 thermodynamic temperatures.

2) VB = 61200

1200

+T

R

3.6 1.2 = 61200

1200

+T

R

0.8 =T

R+12001200

OR 0.4 =T

R+12001200

RT = 300 OR RT = 1800

T = 325 K OR T = 289 K

Hence, the lower temperature will be 289 K.

17.I1 + I2 = I3

I2I3 14 + 10 = 4I2 + 6I3

X10 = 6I3 + 2I1

I1Solving, I1 = 1 A

I2 = 3A

I3 = IXY = 2 A

VXY = 10 - 6 (IXY) = 2 V

Note: CurrentI1 is actually flowing in the direction opposite to the assumed directionin the diagram.

Note: The potential at Y is 2 V lower than the potential at X.


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Assignment Answer

1) The figure on the right shows a circuit containing a30 V battery and 6 resistors. The potential differences

across A, C and D are 22 V, 8 V and 12 Vrespectively. Find the potential difference acrosseach of the components B, E and F and also thepotential at the points U, W, X, Y and Z. (Ans: 8 V, 10V, 10 V, 0 V, 12 V, 20 V, -2 V, -10 V)Sketch a graph to show how the potential variesalong the line XZ. Label the graph with appropriatevalues.

Solution

VYZ = 30 22 = 8 V i.e p.d. across B is 8 V

VUZ = 30 8 12 = 10 V i.e. p.d. across E and F are 10 V.

VU = 0 V (since it is connected to earth)

VW = 0 + 12 = 12 V

VX = 12 + 8 = 20 V

VY = 20 22 = -2 V

VZ = -2 - 8 = -10 V

30 V

C

A B

D

E

F

YZX

WU

V/V

-2 Distance

20

-10

X Y Z


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2) In the figure, AB is a 10 slide wire, 50 cm long. E1 is a 2 V accumulator ofnegligible resistance. R1 and R2 are resistances of 15 and 5 respectively.When the keys K1 and K2 are both open, the galvanometer shows no deflectionwhen AJ is 31.25 cm. When keys K1 and K2 are both closed, the balance length AJ

is 5 cm. Calculate

i. The e.m.f. of cell E2

ii. The internal resistance of the cell E2

iii. The balance length AJ when K2 is open and K1 is close.

iv. The balance length AJ when K2 is close and K1 is open.

(Ans: 0.5 V, 7.5 , 12.5 cm, 12.5 cm)

Solution

(i) When K1 is opened, VAB =10

(2)10 15+

= 0.8 V

Since K2 is opened, no current flows in the lower circuit i.e. e.m.f. E2 is measured.

E2 = VAJ =31.25

(0.8)

50

= 0.5 V

(ii) When K1 is closed, VAB = 2 V

Since K2 is closed, current flows in the lower circuit VAJ = E2 Ir.

VAJ =5

(2)50

= 0.2 V

Since IR = 0.2 V I (5) = 0.2 I = 0.2/5 = 0.04 A

A BJ

E1

E2

R1

R2

K1

K2

15

5


8/8

E2 Ir = 0.2 0.5 0.04r = 0.2 r = 7.5

(iii) When K2 opened, VAJ = E2 = 0.5 V

When K1 is closed, VAB = 2 V

0.5

50 2

l= l = 12.5 cm

(iii) When K2 closed, VAJ = 0.2 V

When K1 is opened, VAB = 0.8 V

0.2

50 0.8

l= l = 12.5 cm

*** End of Tutorial and

Assignment Solution ***

2009 direct current circuits tutorial revised solutions

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