1. (d) 25
2. (b) IR2
3. (d) 25 W
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 (p) 1.6 × 10-8 elm, resistance (R) = 10 Ω.
R = ρL/A
L = πD2R/4p
= 22 × (5 × 10-4)2/ 7 × 4 × 1.6 × 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 as shown below:
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 × 10-3 A
Resistance (R) = V/I = 12V/ 2.5 × 10-3 A = 4800
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 CI 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/nΩ
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.
11. It is given here that R1 = R2 = 611.
(i) To get net resistance of 9 Ω we should join there resistors as below:
(ii) To get 4Ω net resistance we should join three resistors as below:
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 × 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 = 12 Ω
Current (I) = V/R = 220V / 1211 = 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/ la = 2 A/.
Power used in 211 resistor = 12R = 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 +12 = 3/11+ 5/11 = 8/11 = 0.73 A.
16. Energy used by a TV set of power 250 Win 1 hour = P x t = 250 Wh.
Energy used by toaster of power 1200 Win 10 minute (10/60 h)
= P × t = 1200 Wx 10/60 h = 200 Wh.
17. Resistance of electric heater (R) = 8Ω, current (I) = 15 A.
Rate at which heat developed in the heater = I2Rt/t = 15 × 15 × 8 = 1800 W.
18. (a) For filament of electric lamp we require a strong metal with high melting
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.
19. An electric circuit is a continuous and closed path of an electric current. If the electric circuit complete, current can flow through the circuit.
20. SI unit of electric current is Ampere. Current is said to be 1 ampere, if 1 coulomb charge flows per second across a cross-section of conductor.
21. Charge on one electron = 1.6 × 10-19 coulomb. No of electron in one coulomb of charge = 1/1.6 × 10-19 = 6.25 × 1018
23. The potential difference between two points is said to be 1 volt, if 1 joule of work is to be done for moving charge of 1 coulomb from one point to another.
24. Potential difference (V) = 6 V
Charge (Q) = 1 C
Energy = total work done (VV) = Q × V = 1 × 6 = 6 joule.
25. The resistance of a conductor depends on
i. length of conductor (I)
ii. Area of cross-section (A)
iv. Nature of material used to make conductor
26. The current flows more easily through a thick wire as compared to thin wire of the same material, when connected to the same source. It is due to the reason that resistance increases with decrease in thickness.
27. It is given that resistance R of the electrical component remains constant but the potential difference across the ends of the component decreases to half of its value. Hence, as per Ohm’s law, new current also decreases to half of its original value.
28. Coils of electric toasters and electric irons are made of an alloy due to the following reasons:
i. Resistivity of an alloy is generally higher than that of pure metal.
ii. At high temperature, an alloy does not oxidize readily. Hence, coil of an alloy has longer life.
29. (a) Iron is the better conductor than mercury.
(b) Silver is the best conductor because its resistivity is the least.
30. The schematic diagram of circuit is as follows:
Here ammeter A has been joined in series of circuit and voltmeter V is joined in parallel to 12 ohms resistor.
Total voltage of battery V = 3 × 2 = 6 V.
Total resistance R = R1 + R2 + R3 = 5 Ω + 8 Ω + 12 Ω = 25 Ω
Ammeter reading (current) = I = V/R = 6/25 = 0.24 A.
Voltmeter reading = IR = 0.24 × 12 = 2.88 V.
32. When the resistances are joined in parallel, the resultant resistance in parallel arrangement is given by:
1/R = 1/R1 + 1/R2 + 1/R3
(a) 1/R = 1/1+ 1/106 = 1+ 10-6
R = 1Ω
(b) 1/R = 1/1 + 1/103 + 1/106 = 1 + 10-3 + 10-6
R = 1
33. Here, voltage (V) = 220V
R1 = 100Ω, R2 = 50Ω and R3 = 500Ω 1/R = 1/R1 + 1/R2 + 1/R3
1/R = 1/100 + 1/50 + 1/500 = 16/500
R = 500/16 = 31.25Ω
The resistance of electric iron, which draws as much current as all three appliances take together = R = 31.25Ω.
Current passing through electric iron (I) = V/R = 220/31.25 = 7.04A.
34. Advantage of connecting electrical devices in parallel with the battery are as follows:
(i) Voltage across each connecting electrical device is same and device take current as per its resistance.
(ii) Separate on/off switches can be applied across each device.
(iii) Total resistance in parallel circuit decreases, hence, a great current may be drawn from cell.
(v) If one electrical device is damaged, then other devices continue to work properly.
35. (a) If we connect resistance of 3Ω and 6Ω in parallel and then resistance of 2Ω is connected in series of the combination, then total resistance of combination is 4Ω.
(b) If all the three resistance are joined in parallel the resultant resistance will be 3Ω.
36. (a) To obtain highest resistance, all the four resistances must be connected in series arrangement. In that case resultant R = R1 + R2 + R3
= 4 + 8 + 12 48Ω
(b) To obtain lowest resistance, all the four resistance must be connected in parallel 1/R = 1/R1 + 1/R2 + 1/R3
= 1/4 +1/8 +1/12 + 1/24 = 12/24Ω
= 24/12 = 2Ω
37. Cord of heater and electric heater are joined in series and carry same current when joined to voltage source. As resistance of cord is extremely small as compared to that of heater element. hence, heat produced is extremely small in cord but much larger in heater element. So, the heating element begins to glow but cord does not glow.
38. Charge transferred (Q) = 96000 C, time = 1 hour = 60 × 60 = 3600 s and potential difference (V) = 50 V.
Heat generated (H) = VIt = V.Q = 50 x 96000 = 4800000 j = 4.8 × 106 j.
39. Resistance of electric iron (R) = 20Ω, current (I) = 5 A and time = 30 s.
Heat generated (H) = I2Rt = 52 × 20 × 30 = 15000j.
40. Electric power determines the rate at which energy is delivered by a current.
41. It is given that current drawn by electric motor (I) = 5 A. the line voltage V = 220 V time (t) = 2 h.
Power of motor (P) = P = VI = 220 × 5 = 1100 W and the energy consumed (E) = Pt
1100 W × 2 h = 2200 Wh or 2.2kWh.