Question 1 State Joule’s law of heating. list two special characteristics of a heating element wire.
An electronic iron consumed energy at the rate of 880 W when heating is at the maximum rate and 440 W when the heating is at the minimum rate. The applied voltage is 220 V. Calculate the current and resistance in each case Solution
We know, $P = VI$
And $I= \frac {V}{R}$
So
$P= \frac {V^2}{R}$
$R = \frac {V^2}{P}$
Case Maximum Heating
At maximum heating, power is P = 840 W
Working voltage is 220 V
So, resistance at maximum heating is = (220)2/840 = 57.6 Ohm
Now I = P/V
So, current at maximum heating is = 840/220 = 3.8 A Case Minimum Heating
At minimum heating, power is = 360 W
The resistance at minimum heating will be = (220)2/360 = 134.4 Ohm
The current at minimum heating will be = 360/220 = 1.6 A
Question 2 Copper and aluminium wires are usually employed for electricity transmission Why? Solution
Copper and aluminium wires are usually employed for electricity transmission for the below reasons
1)These are extremely good conductors
2) They have a low value of resistivity.So while transmission over long distances, electricity is not dissipated in form of heat and energy is, hence, not wasted.
3)these are ductile and can be drawn in the form of fine wire
4) They are less expensive than the best conductors, silver and gold.
5)They are extremely flexible and do not degrade under electrical transmission as do other metals.
Question 3 Obtain the expression for the heat developed in a resistor by the passage of electric current through it. 220 J of heat is produced each second in a 8 ohm resistor. Find the potential difference across the resistor. Solution
i) Let t be the time during which charge Q flows. Now when charge Q moves against the potential difference V , then the amount of work is given by
Therefore the source must supply energy equal to VQ in time t. Hence power input to the circuit by the source is
The energy supplied to the circuit by the source in time t is P×t that is, VIt.
This is the amount of energy dissipated in the resistor as heat energy.
Thus for a steady current I flowing in the circuit for time t , the heat produced is given by
Applying Ohm's law to above equation we get
This is known as Joule's Law of heating
ii) Now H=V2t/R
Here H=220J t=1 sec R=8 ohm
V2=1760
V=42 Volt
Question 4 (a) What is heating effect of current? List two electrical appliances which work on this effect.
(b) An electric bulb is connected to a 220 V generator. If the current drawn by the bulb is 0.50 A; find its power.
(c) An electric refrigerator rated 400 W operates eight hours a day. Calculate the energy per day in kWh.
(d) State the difference between kilowatt and kilowatt hour.
e) Why is the series arrangement of appliances not used for domestic circuits? Solution
a) When electric current passes through a high resistance wire, the wire becomes and produces heat. This is called heating effect of current. electric Iron,Electric Heater are two appliances which work on this effect
b) P=VI
=220X.50=110W
c) Usage per day= 400 x 8 = 3200 Whr =3.2 KWh
d) Kilowatt is measure of Power while Kilowatt hour is the measure of energy
Question 7 How much energy is given to each coulomb of charge passing through a 6 V battery? Solution
Work done = Potential difference × charge
Here P.D=6V and Q=1C
So work done=6J
Question 8 A copper wire has a diameter 0.5 mm and resistivity 1.6 Â 10-8ohm m.
(i) What will be the length of this wire to make the resistance of 12 Ohm
(ii) How much will be the resistance of another copper wire of same length but half the diameter? Solution
a) Resistance of the wire is given by
$R=\rho \frac{l}{A}$
Here R=12 ohm,resistivity =1.6 X 10-8ohm m. And $A =\pi (d/2) ^2=\pi (.0005/2)^2$
So
$l= \frac{RA}{\rho}$
Substituting the above values
l=147.18 m
b) Now when diameter is halved ,length remains same
$A =\pi (d/2) ^2=\pi (.00025/2)^2$ and l=147.18 m
$R=\rho \frac{l}{A}$
Substituting all the values we get
R=48 ohm
Question 9 A wire of uniform cross-section and length l has a resistance of 4 ohm. The wire is cut into four equal pieces. each piece is then stretched to length ‘l’. Thereafter, the four wires are joined in parallel. Calculate the net resistance Solution
Let the resistivity of the material is ?, length of the wire initially l and the area of cross section is A. Then the resistance of the wire will be,
$R=\rho \frac{l}{A}=4$
Now cut the wire in equal four pieces and stretch it to the initial length l. When we stretch it the new length will be same of each piece but the area get reduced to one fourth ,So resistance of that piece would become
$R_1=\rho \frac{l}{A/4}=4\times 4=16$
Now, connect these four pieces in parallel. Then the equivalent resistance will be,
$\frac {1}{R_equiv}=\frac {1}{R_1} +\frac {1}{R_2}+ \frac {1}{R_3}+\frac {1}{R_4}$
$\frac {1}{R_equiv}=\frac {1}{16} +\frac {1}{16}+ \frac {1}{16}+\frac {1}{16}$
Requiv=4 ohm
Question 10 Calculate the electrical energy produced in 5 minutes when a current of 2 A is sent through a conductor by a potential difference of 500 volts. Solution
$P = V \times I = 2 \times 500 =1000W$
$E= P \times t= 1000 \times 300 = 3 \times 10^5$ J
Question 11 An electric heater draws a current of 10 A from a 220 V supply. What is the cost of using the heater 5 hours per day for 30 days if the cost of 1 unit is Rs. 2.50? Solution
$P=VI$
$= 220 \times 10$
= 2200W
Electricity consumption in a day
$E=P \times T$
$=2200 \times 5$#
= 11000W=11 KWH /day
Usage per month=11*30 = 330KWH
Cost of the Electricity= 330* 2.50= RS.825
Question 12 An electric bulb draws a current of 0.2 A when it operates at 220 V. Calculate the amount of electric charge flowing through it in 1 h Solution
I=.2 A
V=220 V
Now
t= 1 hour = 3600 sec
Now
$I = \frac {q}{t}$
or
$q = I \times t = .2 \times 3600 = 720 C$
Question 13 An electric bulb is rated 220 V and 100 W. when it is operated on 110 V, what will be the power consumed? Solution
Question 14 Resistors are given as R1 = 10 ohm, R2 = 20 ohm, and R3 = 30 ohm. Calculate the effective resistance when they are connected in series. Also calculate the current flowing when the combination is connected to a 6 V battery. Solution
Given R1 = 10 ohm, R2 = 20 ohm, and R3 = 30 ohm
$R=R_1 + R_2 + R_3 = 10 +20 + 30 = 60 /omega$
Now from Ohm's Law
$V=IR$
$I = \frac {V}{R} = .1 A$
Question 15 The filament of an electric lamp, which draws a current of 0.2 A, is used for 5 hours. Calculate the amount of charge flowing through the circuit. Solution
I=.2 A
Now
t= 5 hour = 18000 sec
Now
$I = \frac {q}{t}$
or
$q = I \times t = .2 \times 18000 = 3600 C$
Question 16 Calculate the number of electrons passing per second through a conductor to produce a current of one ampere. Solution
1 Ampere means 1 C/sec.So we need to find the number of electrons in 1 C
Now Charge on 1 electron=$1.6 \times 10^{-19} C$
So Number of electrons on 1 C
$= \frac {1}{1.6 \times 10^{-19}} =6.25 \times 10^{18}$
Question 17 A 5 ohm resistor is connected across a battery of 6 volts. Calculate
(i) The current flowing through the resistor.
(ii) The energy that dissipates as heat in 10 s. Solution
R=5 ohm, V= 6 Volt, I=?
$I = \frac {V}{R} = 1.2 A$
Now Energy dissipated
H= VIt= 300 J
Question 18 You have two electric lamps having rating 40 W; 220 V and 60 W; 220 V. Which of the two has a higher resistance? Give reason for your answer. If these two lamps are connected to a source of 220 V, which will glow brighter? Solution
Power is given by
$P = \frac {V^2}{R}$
So, resistance can be obtained as
$R = \frac {V^2}{P}$
For Electric Lamp A
$R = \frac {V^2}{P} =\frac {220^2}{40} =1210 \omega $
For Electric Lamp B
$R = \frac {V^2}{P} =\frac {220^2}{60} =806.66 \omega $
So, Lamp A has higher Resistance
Now Lamp B will glow brighter when connected with 220 V as the power is higher than lamp A
Question 19 A current of 5 ampere is passed through a conductor of 12 ohms for 2 minutes. Calculate the amount of heat produced. Solution
Question 20) Can you run an electric geyser with power rating 2 kW; 220 V on a 5 A line? Give reason to justify your answer. Solution
$P=VI$
or $I= \frac {P}{V} = 9.09 A$
Thus, the electric geyser draws current much more than the current rating of 5 A
So, we cannot run this electric geyser on the 5 A line
Question 21 A domestic electric circuit (220 V) has a 5 A fuse. How many bulbs of 100 W; 220 V rating can be safely used on this line? Solution
$P=\frac {V^2}{R}$
$R=\frac {V^2}{P}$
For one bulb, resistance would be
$R=\frac {220^2}{100}=484 \omega$
Current will be given as
I=V/R = 220/484 =.45 A
Now One bulb drawing the 0.45 amps
11 bulbs drawing the 4.95 amps (<5 amps)
So, the number of 100 watts bulbs to connect with out blowing the fuse are 11.
Question 22 Two bulbs A and B are rated as 90 W -120V and 60 W -120 V respectively. They are connected in parallel across a 120 V source. Find the current in each bulb. Which bulb will consume more energy?
Solution
$P =VI$
or $I = \frac {P}{V}$
Fore bulb A
I=.75 A
Fore bulb B
I=.5 A
Higher Power rating bulb will consume more energy, So Bulb A
Question 23 An electric iron is rated 2 kW at 220 V. Calculate the capacity of the fuse that should be used for the electric iron. Solution
$P =VI$
or $I = \frac {P}{V}$
I= 2000/220 = 9.09 A
Thus,10 A fuse should be used with electric Iron
Question 24) For an electric heater rated 4 kW-220 V. Calculate:
(i) the current required
(ii) the resistance of the heater
(iii) the energy consumed in 1 hour. Solution
$P =VI$
or $I = \frac {P}{V}$
I= 4000/220 = 18.18 A
$P =\frac {V^2}{I}$
or $R = \frac {V^2}{P}=12.1 \omega$
Now Energy consumed in 1 hour
= P X time = 4KWH
