"AKC SCIENCE CLASSES"
CLASS 10 TH (CBSE AND MP BOARD)
CHAPTER 12
ELECTRICITY
EXERCISES QUESTIONS AND ANSWERS
Q.01:- 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
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Ans:- (d) Resistance of each parts = R/5
When the five parts are connected in parallel, the equivalent resistance R' is given by
1/R' = 5/R + 5/R + 5/R + 5/R + 5/R = 25/R
R/R' = 25. ANS.
Q.02:- Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Ans:- (b) IR² does not represent electrical power.
Q.03:- An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Ans:- (d) Resistance,
R = V²/P = (200)²/100 = 484 Ω
When operated on 110V, the power consumed will be
P' = V'²/R = (110)²/484 = 25 W. Ans.
Q.04:- Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference.
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Or
Two identical wires are first connected in series and then in parallel to a source of supply. Find the ratio of the heat produced in two cases.
Ans:- (c) 1 : 4
Let R be the resistance of each wire. In series combination, the total resistance will be 2R. Heat produced,
Hs = V²t/2R
In parallel combination, the total resistance is R/2. Heat produced,
Hp = V²t/(R/2) = 2V²t/R = 4 Hs or Hs/Hp = 1/4. Ans.
Q.05:- How is voltmeter connected in the circuit to measure the potential difference between two points?
Ans:- A voltmeter is connected in parallel to measure the potential difference between two points in a circuit.
Q.06:- 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?
Ans:- Radius, r = 0.5/2 = 0.25 mm = 0.025 cm
ρ = 1.6 ✖ 10⁻⁶ Ω cm, R = 10 Ω, L = ?
As R = ρ ------- = ρ -------
A πr²
L = -------- = ----------------------------------
ρ 1.6 ✖ 10⁻⁶
= 12265.625 cm ⋍ 122.6 m. Ans.
Thus, when the diameter of wire is doubled, the resistance becomes one-fourth of the original value.
New resistance = 10/4 = 2.5 Ω
Decrease in resistance = 10 - 2.5 = 7.5 Ω. Ans.
Q.07:- The value of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below :-
Plot a graph between V and I and calculate the resistance of the resistor.
Ans:- The graph between V and I for the data is shown below :-
As R = ----------- = --------------- = -----------
Q.08:- 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.
Ans:- Here, V = 12 V, I = 2.5 mA = 2.5 ✖ 10⁻³ A
Resistance, R = V/I = 12 V/ (2.5 ✖ 10⁻³ A) = 4800 Ω. Ans.
Q.09:- A battery of 9V 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?
Ans:- Total resistance,
R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
Potential difference, V = 9 V
Current through 12 Ω resistor = 0.67 A. Ans.
Q.10:- How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Ans:- Suppose n resistance of 176 Ω are connected in parallel. Then
1/R = 1/176 + 1/176 = ....... n factors = n/176
Or R = 176/n Ω.
By Ohm's law,
R = V/I
176/n = 220/5 or n = (176 ✖ 5)/220 = 4. Ans.
Q.11:- Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of
(a) 9 Ω
(b) 4 Ω
Ans:- Here R₁ = R₂ = R₃ = 6 Ω
(a) When we connect R₁ in series with the parallel combination of R₂ and R₃, as shown in figure (a).
The equivalent resistance is
R = R₁ ------------- = 6 + --------------
R₂ + R₃ 6 + 6
= 6 + 3 = 9 Ω. Ans.
(b) When we connect a series combination of R₁ and R₂ IN parallel with R₃, as shown in figure (b) the equivalent resistance is
R = ------------- = -------- = 4 Ω. Ans.
12 + 6 18
Q.12:- Several electric bulbs designed to be used on a 220V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A ?
Ans:- Resistance of each bulb, R = V²/P = (220)²/10 = 4840 Ω.
Suppose n bulbs are needed to be connected in parallel with each other. Then their equivalent resistance Rp is given by
1/Rp = 1/4840 + 1/4840 + .......n factors = n/4840
Or Rp = 4840/n Ω
Given, V = 220 V, I = 5 A
By Ohm's law, Rp = V/I Or 4840/n = 220/5
or n = (4840 ✖ 5)/220 = 110. Ans.
Q.13:- A hot plate of an electric oven connected to a 220 V line has resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or parallel. What are the currents in the three cases?
Ans:- (ⅰ) When the two coils A and B are used separately,
R = 24 Ω, V = 220 V
Current, I = V/R = 220V/24Ω = 9.167 A. Ans.
(ⅱ) When the two coils are connected in series,
R = 24 + 24 = 48 Ω, V = 220 V
Current, I = V/R = 220/48 = 4.58 A. Ans.
(ⅲ) When the two coils are connected in parallel,
R = (24 ✖ 24)/(24+24) = 12 Ω, V = 220 V
Current, I = V/R = 220/12 = 18.33 A. Ans.
Q.14:- Compare the power used in the 2 Ω resistor in each of the following circuits :-
(ⅰ) a 6 V battery in series with 1 Ω and 2 Ω resistors and,
(ⅱ) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Ans:- (ⅰ) The circuit diagram is shown in figure.
Potential difference, V = 6 V
Current, I = V/R = 6V/3Ω = 2 A
Power used in 2 Ω resistor = I²R = (2)² ✖ 2 = 8W. Ans.
(ⅱ) The circuit diagram for this case is shown below :-
Q.15:- Two lamps one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric main supply. What current is drawn from the line if the supply voltage is 220 V?
Ans:- Total power consumed in the circuit = 100 + 60 = 160 W
Voltage, V = 220 V
But, Power = VI
ஃ Current, I = Power/V = 160/220 = 0.727 A. Ans.
Q.16:- Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
= 250 W ✖ 1 h = 250 Wh
= 1200 W ✖ 10 min = 1200 W ✖ 10/60 h = 200 Wh
Q.17:- An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Ans:- Here R = 8 Ω, I = 15 A, t = 2 h
The rate at which heat is developed in the heater is equal to the power.
ஃ P = I²R = (15)² ✖ 8 = 1800 Js⁻¹. Ans.
Q.18:- Explain the following :-
(a) Why is the tungsten used almost exclusively for filament of electric lamps (bulbs)?
(b) Why are the conductors (coils) of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metals?
(c) Why is the series arrangement not used for domestic circuits ?
Or
Series arrangements are not used for domestic circuits. Give reasons.
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Ans:- (a) This is because thin wire of tungsten has high resistance and high melting point (3410℃). When a current is passed through it, it becomes hot and emits light.
- As compared to pure metals, alloys have higher resistivity. When a current is passed through the element made of alloy, a large amount of heat is produced.
- Resistivity of an alloy almost does not change with temperature. So an element of alloy does not readily get oxidised at high temperature.
- In series arrangement same current will flow through all the appliances, which is not desired.
- Total resistance (= sum of the resistances of all appliances) becomes large, and the current gets reduced.
- We cannot use independent ON/OFF switches with individual appliances.
- All appliances have to be used simultaneously even if we don't need them.
