0
1. (d) 25

2. (b) IR2

3. (d) 25 W

4. (c)1:4

5. A voltmeter is always connected in parallel to resistance across the point between
which the potential difference is to be measured.

6. Diameter of wire (d) = 0.5 mm, resistivity (ρ) 1.6 x 10-8 Ωm, resistance (R) = 10 Ω.
 R = ρL/A
 L= πD2R/4ρ
 = 22 x (5 x 10-4)
2/ 7 x 4 x 1.6 x 10-8 = 122.5 m
If the diameter is doubled for given length of given material resistance is inversely
proportional to the cross-section area of wire.

7. From the given data the I-V graph is a straight line

Resistance of resistor (R) = VA-VB/1A-1B = 12 V – 6 V/ 3.6 A – 1.8 A
 = 6V/ 1.8 A = 3.3 Ω

8. Voltage of battery = V = 12 V, Current ( I) = 2.5 mA = 2.5 x 10-3 A
 Resistance (R) = V/I = 12V/ 2.5 x 10-3 A = 4800 Ω.

9. Potential difference (V) = 9 V.
 Total resistance (R) = R1+ R2+ R3+R4 +R5
 = 0.2 +0.3 + 0.5 + 0.5 + 12 = 13.4 Ω
 Current in the circuit (I) = V/R = 9 V / 13. 4 Ω = 0.67 A.
In series circuit same current flows through all the resistance, hence current of 0.67
A will flow through 12 Ω resistor.

10. Let a resistors of 176Ω are joined in parallel. Then their combined resistance (R)
 1/R = 1/176 + 1/176 …… times = n/176 or R = 176/n Ω
 It is given that V= 220 V and I = 5 A
 R = V/I or 176/n = 220/5 = 44Ω
n = 176/44 = 4, 4 resistors should e joined in parallel.

12. Each bulb is rated as 10 W, 220 V, It draws a current (I) = P/V = 10W/220 V
 = 1/22 A.
As the maximum allowable current is 5A and all lamps are connected in parallel,
hence maximum number of bulbs joined in parallel with each other = 5 x 22 = 110.

13. It is given that potential difference (V) = 220 V.
 Resistance of coil (A) = Resistance of coil (B) = 24 Ω
(i) When either coil is used separately, the circuit (I) = V/R = 220 V/ 24 Ω
= 9.2 A.
(ii) When two coils are used in series total resistance (R)
 = R1 + R2 = 24 +24 = 48Ω
Current flowing (I) = V/ R = 220 V/ 48 Ω = 4.6 A.
(iii) When two coils are joined in parallel. Total resistance (R) = 1/24 + 1/24
= 2/24, R = 12Ω.
Current (I) = V/R = 220V / 12Ω = 18.3 A.

14. (i) When a 2Ω resistor is joined t a 6 V battery in series with 1Ω and 2Ω resistors.
Total resistance (R) = 2 + 1 + 2 = 5Ω.
Current (I) = 6V/5Ω = 1.2 A
Power used in 2 A resistor = I2R = 2.88 W
(ii) When 2Ω resistor is joined to a 4 V battery in parallel with 12Ω resistor and 2Ω
resistors, the current flowing in 2Ω = 4 V/ 2Ω = 2 A/.
Power used in 2Ω resistor = I2R = 8 W
Ratio = 2.88/8 = 0.36: 1.

15. Current drawn by 1st lamp rated 100 W at 220 V = P/V = 100/ 220 = 5/11 A.
 Current drawn by 2nd lamp rated 60 W at 220 V = 60/220 = 3/11 A.
 In parallel arrangement the total current = I1 +I2 = 3/11+ 5/11 = 8/11 = 0.73 A.

16. Energy used by a TV set of power 250 W in 1 hour = P x t = 250 Wh.
 Energy used by toaster of power 1200 W in 10 minute (10/60 h)
 = P x t = 1200 W x 10/60 h = 200 Wh. Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in
Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks

17. Resistance of electric heater (R) = 8Ω, current (I) = 15 A.
 Rate at which heat developed in the heater = I2Rt/t = 15 x 15 x 8 = 1800 W.

18. (a) For filament of electric lamp we require a strong metal with high melting
point. Tungsten is used exclusively for filament of electric lamps because its
melting point is extremely high.
(b) Conductors of electric heating devices are made of an alloy rather than a pure
metal due to high resistivity than pure metal and high melting point to avoid
getting oxidized at high temperature.
(c) Series arrangement is not used for domestic circuits as current to all
appliances remain same in spite of different resistance and every appliance
can not be switched on/ off independently.
(d) Resistance of a wire is inversely proportional to its cross-section area.
(e) Copper and aluminium wires are usually employed for electricity
transmission because they are good conductor with low resistivity. They are
ductile also to be drawn into thin wires.

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