Science Class- 10
Class 10 - Chapter 11: Electricity
Q1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R′/R is
Answer: (a) 1/25
When a wire of resistance R is cut into 5 equal parts, resistance of each part becomes R/5.
These 5 equal resistors are connected in parallel.
Equivalent resistance,
R′ = (R/5)/5 = R/25
Therefore,
R′/R = 1/25
Q2. Which of the following terms does not represent electrical power in a circuit?
Answer: (a) I²R
Electrical power is given by:
- P = VI
- P = V²/R
- P = I²R
I²R represents power, so there appears to be a printing error in the options. The correct power expressions are VI, V²/R and I²R.
Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be
Answer: (d) 25 W
Given:
V = 220 V, P = 100 W
Resistance of bulb,
R = V²/P = 220²/100 = 484 Ω
When operated at 110 V,
P = V²/R
P = 110²/484
P = 25 W
Q4. Two conducting wires of the same material and equal lengths and equal diameters are first connected in series and then in parallel across the same potential difference. The ratio of heat produced in series and parallel combinations would be
Answer: (a) 1 : 2
Let resistance of each wire be R.
Series combination:
Rs = 2R
Parallel combination:
Rp = R/2
Heat produced H ∝ 1/R
Therefore,
Hs : Hp = (1/2R) : (2/R)
= 1 : 4
Answer: (c) 1 : 4
Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
A voltmeter is connected in parallel across the two points between which the potential difference is to be measured.
It has very high resistance so that it draws negligible current from the circuit.
Q6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Given:
| Quantity | Value |
|---|---|
| ρ | 1.6 × 10-8 Ω m |
| d | 0.5 mm |
| R | 10 Ω |
Area,
A = πd²/4
A = 1.96 × 10-7 m²
Using R = ρL/A
L = RA/ρ
L = (10 × 1.96 × 10-7)/(1.6 × 10-8)
L = 122.5 m
Length of wire = 122.5 m
If diameter is doubled, area becomes four times.
Since resistance is inversely proportional to area, resistance becomes one-fourth.
New resistance = 10/4 = 2.5 Ω
Q7. Plot a graph between V and I and calculate the resistance of that resistor.
Given Data:
| I (A) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |
|---|---|---|---|---|---|
| V (V) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |
The graph between V and I is approximately a straight line passing through the origin.
Resistance = Slope of V-I graph
R = V/I
Using V = 13.2 V and I = 4 A
R = 13.2/4
R = 3.3 Ω
Q8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Given:
V = 12 V
I = 2.5 mA = 0.0025 A
Using Ohm's law,
R = V/I
R = 12/0.0025
R = 4800 Ω
Resistance = 4.8 kΩ
Q9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω respectively. How much current would flow through the 12 Ω resistor?
Total resistance,
R = 0.2 + 0.3 + 0.4 + 0.5 + 12
R = 13.4 Ω
Current,
I = V/R
I = 9/13.4
I = 0.67 A
Since resistors are connected in series, the same current flows through every resistor.
Current through 12 Ω resistor = 0.67 A
Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Total current = 5 A
Voltage = 220 V
Equivalent resistance required,
R = V/I
R = 220/5
R = 44 Ω
For n identical resistors in parallel,
R = 176/n
44 = 176/n
n = 4
Answer: 4 resistors
Q11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
(i) To obtain 9 Ω:
Connect two 6 Ω resistors in parallel.
Equivalent resistance = 3 Ω
Connect this combination in series with the third 6 Ω resistor.
Total resistance = 3 + 6 = 9 Ω
(ii) To obtain 4 Ω:
Connect two 6 Ω resistors in series.
Equivalent resistance = 12 Ω
Connect this combination in parallel with the third 6 Ω resistor.
Equivalent resistance = 4 Ω
Q12. Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel across the two wires of 220 V line if the maximum allowable current is 5 A?
Current drawn by one bulb,
I = P/V
I = 10/220
I = 0.045 A
Maximum allowable current = 5 A
Number of bulbs,
n = 5/0.045
n ≈ 110
Answer: 110 bulbs
Q13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance.
(i) Used separately:
I = V/R = 220/24 = 9.17 A
(ii) Used in series:
R = 24 + 24 = 48 Ω
I = 220/48 = 4.58 A
(iii) Used in parallel:
R = 12 Ω
I = 220/12 = 18.33 A
Q14. Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) 6 V battery with 1 Ω and 2 Ω resistors in series
Total resistance = 3 Ω
Current = 6/3 = 2 A
Power in 2 Ω resistor:
P = I²R
P = 2² × 2
P = 8 W
(ii) 4 V battery with 12 Ω and 2 Ω resistors in parallel
Voltage across 2 Ω resistor = 4 V
P = V²/R
P = 16/2
P = 8 W
Comparison: Power consumed in both cases is 8 W.
Q15. Two lamps, one rated 100 W at 220 V and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Total power = 100 + 60 = 160 W
Current drawn,
I = P/V
I = 160/220
I = 0.73 A
Answer: 0.73 A
Q16. Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?
Energy used by TV:
E = 250 × 1 = 250 Wh
Energy used by toaster:
E = 1200 × (10/60)
E = 200 Wh
Answer: The 250 W TV used for 1 hour consumes more energy.
Q17. An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Rate of heat production = Power
P = I²R
P = 5² × 44
P = 1100 W
Answer: Heat is developed at the rate of 1100 W.
Q18(a). Why is tungsten used almost exclusively for filament of electric lamps?
Tungsten has a very high melting point (about 3380°C) and high resistivity. It can therefore be heated to very high temperatures without melting and emits bright light.
Q18(b). Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Alloys have higher resistivity and higher melting points than pure metals. They do not oxidise easily at high temperatures and are therefore suitable for heating elements.
Q18(c). Why is the series arrangement not used for domestic circuits?
In a series circuit:
- The same current flows through all appliances.
- If one appliance fails, the entire circuit breaks.
- Appliances cannot be operated independently.
Therefore, series arrangement is not suitable for domestic wiring.
Q18(d). How does the resistance of a wire vary with its area of cross-section?
Resistance is inversely proportional to the area of cross-section.
R ∝ 1/A
Thus, increasing the area decreases the resistance.
Q18(e). Why are copper and aluminium wires usually employed for electricity transmission?
Copper and aluminium have very low resistivity and are good conductors of electricity. They allow electric current to flow with minimum loss of energy and are therefore used for transmission lines.