|
Electric
Current (I) |
I = Q/t, SI unit: Ampere |
|
Potential Difference (V) |
V = W/Q, SI unit: Volt |
|
Ohms Law |
V = IR, R is Resistance, Unit: Ohms |
|
Resistivity (ρ) |
ρ= R.A/L, Unit: Ohm.meter |
|
Examples of Alloys: |
Nichrome (Ni, Cr, Mn, Fe), Constantin (Cu, Ni),
Manganin (Cu, Mn, Ni), Bronze (Cu, Sn), Brass (Cu, Zn) |
|
Resistance in Series |
Rs = R1 + R2 + R3 ….., The overall resistance increases
in series |
|
Resistance in Parallel |
1/Rp = 1/R1 +
1/R2 + 1/R3 …., The overall resistance decreases in Parallel |
|
Joules law of Heating (H) |
H = I2 R T
or H = V I T |
|
Electric Power (P) |
P = VI or P = I2R or P= V2/R, SI
Unit: Watt |
|
Electric energy Consumed |
E = PxT |
Ans: The current is defined as rate of flow of charge through a conductor.
The SI Unit of electric current is Ampere and is denoted by "A"
When "Q" charge flows for time "T", The current is given by,
I = Q / T
Q. Define potential difference?
Ans: Potential difference is the difference in quantity of electrons across the conductor, resulting in flow of current.
It is defined as work done (W) in moving unit charge (Q) from one point to another.
It is denoted by "V" and its SI unit is Volt.
Q. Describe Ohms Law? Give its experimental verification?
Ans: Ohms law was given by Simon Ohm. It shows relation between Potential (V) across the conductor with Current (I) through it.
Ohms law states that Current flowing in conductor is directly proportional to Potential difference across its ends.
I.e , V ∝ I
V = I × R
Where R is the resistance i.e, Opposition to flow of current.
Experimental Verification:
When the resistance is increased in a circuit, The value of "V" and "I" is measured.
When the graph is plotted, The graph is a straight line.
This is experimental proof of Ohms Law.
Ans: Resistance means the opposition to the flow of current. It is denoted by Ω and its SI unit is Ohm.
The resistance depends on following factors:
1. The resistance increases with increase in length of conductor.
2. The resistance decreases with increase in area of cross section
3. Different materials have different resistance.
4. The resistance increases with increase in temperature.
Q. What is Joules law of heating?
Ans: The Joules law of heating defines the heat produced in a conductor when current flows in it.
It is given by,
2
H = I ✕ R ✕ T
Q. What is Electic Power?
Ans: Electric Power is defined as the rate at which work is done.
It is also defined as rate at which electic energy is consumed.
Power = Work done/ Time
P= W/T
Power is also equal to voltage × current
P = V × I
Its SI unit "Watt"
Q. What do you mean by resistors connected in series?
Ans: When resistors are connected in series, it means they are placed end-to-end along a single path. In series, same current flows through the circuit.
Q. What do you mean by resistors connected in parallel ?
Ans: When resistors are connected in Parallel, it means they are placed side-by- side along a single path. In Parallel, the potential difference across the ends of conductor is same.
Q: Derive the relation for Resistance in Series.
Ans:
Let R1, R2 and R3 be three resistances in series.
Let V1, V2, V3 be the potential difference across R1, R2, R3.
Let "I" be the current flowing through them.
Total Potential difference (V) is,
V = V1 + V2 + V3
But V1 = IR1
V2 = IR2
V3 = IR3
V = IR1 + IR2 + IR3
V = I (R1 + R2 + R3)
V/I = R1 + R2 + R3
R = R1 + R2 + R3
Thus, Total resistance in series is sum of individual resistances.
Q. Derive the relation for resistance in Parallel.
Ans:
Let 3 resistances R1, R2, R3 be connected in Parallel.
Let V be the Potential difference.
Let I1, I2, I3 be current flowing through R1, R2 and R3.
I = I1 + I2 + I3
But, I1 = V/R1,
I2 = V/R2
I3 = V/R3
I.e, I = V/R1 + V/R2 + V/R3
I = V ( 1/R1 + 1/ R2 + 1/ R3)
I/V = 1/R1 + 1/R2 + 1/R3
1/R = 1/R1 + 1/R2 + 1/R3
There total resistance in Parallel is sum of reciprocal of Individual resistances and is less that least.
Q. Will current flow more easily through a thick wire or a thin wire of a same material, when connected to the same source? Why?
Ans: Current will flow more easily through a thick wire because the resistance is inversely proportional to the area of cross-section of the wire. Which means when the area of cross section of the wire increases then resistance will decrease.
Q. When a 12V 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:
V = IR
i.e.; R=V/I
R=12/0.0025
=4800 Ω
Therefore, the resistance of the resistor is 4800 Ω or 4.8 kΩ.
IMPORTANT TEXTUAL QUESTIONS:
Q. Name a device that helps to maintain a potential difference across a conductor.
Ans: Battery helps to maintain a potential difference across a conductor.
Q. What is meant by saying that the potential difference between two points is 1 V?
Ans: When 1 Joule work is done to move a charge of 1 Columb from one point to another, the potential difference between two points is 1 Volt.
Q. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Ans: Because the melting point of an alloy is much higher than a pure metal so they dont melt in high temperatures.
Q. Which among iron and mercury is a better conductor?
Ans: Iron is a better conductor than mercury because the resistivity of mercury is more than the resistivity of iron.
Q. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 106 Ω
Solution: When 1 Ω and 106 are connected in parallel, the equivalent resistance is given by:
1/R = 1/ R1 + 1/ R2
1/R= 1/1 + 1/106
1/R= 106+1/106
1/R= 107/106
R = 106/107
R= 0.99
Therefore, the equivalent resistance is 0.99 Ω.
Q. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Ans: Electrical devices connected in parallel has following advantages:
1. There is no division of voltage among the appliances.
2. If one device fails, Others will work normally.
Q. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Solution:
(a) 3 Ω and 6 Ω are connected in parallel. Hence, their equivalent resistance is given by
1/R= 1/ R1 + 1/R2
1/R = 1/3 + 1/6
1/R= 2+1/6
1/R= 3/6
R= 6/3
R= 2 Ω
Third resistor 2 Ω is connected in series with the 2 Ω resistor, i.e,
R= 2 Ω +2 Ω = 4 Ω
Hence, the total resistance of the circuit is 4 Ω.
(b) All the resistors are connected in parallel. Therefore, their equivalent resistance can be calculated as follows:
1/R = 1/R1 + 1/R2 + 1/R3
1/R = 1/ 2 + 1/3 + 1/6
1/R = 3 + 2 + 1 /6
1/R = 6/6
R = 1
The total resistance of the circuit is 1 Ω.
Q. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Ans:
R = 4Ω + 8 Ω + 12 Ω + 24 Ω
R = 48 Ω.
(b) Lowest resistance will be achieved when coils will be connected in Parallel,
1/ R = 1/4 + 1/8 + 1/12 + 1/24
1/R = 6 + 3 + 2 + 1 /24
1/R = 12 / 24
R = 24 / 12
R = 2 Ω
Hence, the lowest total resistance is 2 Ω.
Q. Why does the cord of an electric heater not glow while the heating element does?
Ans: The heating element of an electric heater is amde of Nichrome.
It has very high resistivity that copper of which cord is made. Therefore it glows.
Q. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
SOLUTION: Here, Charge (Q) : 96000 C
Potential Difference (V): 50 V
Therefore, Heat Generated is 48,00,000 Joules
Q. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
SOLUTION:
According to Joules law of heating,
H = I × I × R x t
H = 5 × 5 × 20 x 30
H = 15, 000 J
The amount of heat developed by the electric iron in 30 s is 15,000 Joules
Q. What determines the rate at which energy is delivered by a current?
Ans: Electric power is the rate at which energy is delivered by a current.
Q. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Ans: The power of the motor can be calculated by the equation,
P = V × I
Substituting the values in the above equation, we get
P = 220 V × 5 A
P = 1100 Watt
The energy consumed by the motor can be calculated using the equation,
E = P × T
Substituting the values in the above equation, we get
P = 1100 W × 7200 S
6
P = 7.92 × 10 J.
The power of the motor is 1100 W and the energy consumed by the motor in 2 hours is
6
7.92 × 10 J.
Ans: Because the melting point and resistance of tungsten is very high.
Due to large resistance, the tungsten glows.
Q. Why are conductors of electric heating devices made of alloy rather than Pure metal?
Ans: Because the resistivity of Alloy is very large than pure metal. So, they do not burn out easily.
Q. Why is series arrangement not used in domestic circuits?
Ans: The equivalent resistance in series is large. It can produce heat and fire.
If one resistance breaks, the whole circuit will not work.
Q: How does the resistance of wire vary with its area of cross section?
Ans: Resistance is Inversely proportional to area of cross section.
The larger the cross section, lower is the resistance.
Q. Why are copper and almunium usually employed for electricity transmission?
Ans: Both copper and Almunium have low resistance, low cost than Silver, hence used in electricity transmission.
Q. If the Potential difference across the ends of conductor decreases to half, what changes will occur in the current through it?
Ans: According to ohm's law, the Potential difference is directly proportional to the Current. If the potential difference gets half then the current is also getting half.
Q. An electric oven of 2 kW power rating is operated in a domes
tic electric circuit that has a current rating of 5A, what result do you
expect? Explain.
Ans: Given; Power = 2kW i.e 2000 W
Voltage = 220V
Current = 5A
We know,
P = V x I
2000 = 220 x I
I = 2000 / 220
I = 9.1 A
The above value exceeds the rating of 5 A for the circuit. Thus the wire will melt and cause short circuit.
Q. An electric lamp of 100 Ω, a toaster of 50 Ω
and a water filter of 500 Ω are connected parallel to a 220 V source.
What is the resistance of the electric iron connected to the same source that takes
as much current as all the three appliances and what is the current through it?
Ans: Given;
Resistance of
electric lamp =100 ohm
Resistance of a toaster = 50 ohm
Resistance of a water filter = 500 ohm
Potential difference = 220 volt
We know,
Resistance in Parallel (1/R) = 1/R1 + 1/R2 + 1/R3
= > 1/R = 1/100 + 1/50 + 1/500
=> 1/R = 5 + 10
+ 1/500
=> 1/R = 16/500 = 4/125
=> R = 125/4
=> R = 31.25 Ω
As given, Resistance of Electric iron is equal to above
value i.e; 31.25 Ω
Thus, Current through the Electic iron =
I = V / R
I = 220 / 31.25
I = 7.04 A
Q. 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’ will be?
Ans: Since Resistance is directly proportional to Length,
Thus resistance of each part = R / 5
Resistance of all 5 parts connected in parallel (R’) =
1 / R’ = 1 + 1 + 1
+ 1
+ 1
R/5 R/5
R/5 R/5
R/5
1 / R’ = 5/R + 5/R
+ 5/R + 5/R + 5/R
1 / R’ =25/R
R/R’ = 25
That means the new resistance will be 25 times less than the original resistance.
Q. An electric bulb is rated 220 V and 100 W, when
operated on 110 V, Calculate the power it will consume?
Ans: Given;
Power, P = 100 W
Potential difference, V = 220 V
We know,
P = V I
P = V x V/R
P = V^2 / R
R = V^2 / P
R = 220 x 220 /
100
R = 484 Ω.
Therefore, for 110 V electric bulb, power consumed will be = V2/R
= > 1102 /484
i.e 25 W.
Q. Two conducting
wires of the same material and of equal lengths and equal diameters are first
connected in series and then in parallel in an electric circuit. Find the ratio
of heat produced in series and parallel combination?
Ans: Let R be the resistance of both wires.
Resistance in series = 2 R
Resistance in parallel = R / 2
Heat produced in Series = I^2 2R T
Heat produced in Parallel = I^2 R/2 T
Ratio of heat produced in Series to Parallel =
Rs / Rp = I^2 2R T / I^2 R/2 T
Rs / Rp = I^2 2R T x 2 / I^2 R T
Rs / Rp = 4 / 1
Thus heat produced in series will be 4 times than heat
produced in parallel
Q. 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 will flow through a 12 Ω resistor ?
Ans: Given,
Potential difference = 9 V
Total resistance in series = 0.2 Ω + 0.3 Ω
+ 0.4 Ω + 0.5 Ω + 12 Ω
Total resistance in series = 13.4 Ω
Since in Series combination, same current flows through
all resistances,
Thus Current, I = V / R
I = 9
/ 13.4
I = 0.67 A
Ans: Given,
R = 176 Ω
I = 5 A
V = 220 V
But resistance (in
parallel) of wire when 5 A current flows when 220 V applied =
R = V / I
R = 220 / 5
R = 44
Therefore, number of 176 Ω resistances that are
required to be connected in parallel for 5A current across 220 V =
176 / 44
= 4
Q. How will you connect three resistors of 6 Ω
each, so that the combination has resistance of i. 9 Ω ii. 4 Ω
Ans:
i. 9 Ω:
To achieve 9 Ω resistance, we need to connect two
in Parallel
i.e; 1/6 + 1/6
= 1 + 1 /6
= 2 / 6
= 6/2
= 3 Ω
Then join third one in series;
i.e; 3 + 6
= 9 Ω
ii. 4 Ω:
To achieve 4 Ω resistance, we need to connect two
in Series
6 + 6 = 12 Ω
Then Join third one in parallel,
1/ 12 + 1 /6
= 1+2/12
=3/12
= 12/3
= 4 Ω
Q. Several electric bulbs designed to be used on a 220 V
electric supply 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 current is 5
A?
Ans:
Given,
Voltage = 220 V
Current = 5 A
Power = 10 W
Thus,
Resistance of each bulb = V2 / P
= 220 x 220 / 10
= 4840 Ω
If ‘n’ number of bulbs of resistance 4840 Ω can be
connected in parallel, the equivalent resistance will be given as;
4840 / n ……………… I
But resistance in parallel R = V / I
= 220 / 5
= 44 Ω ……………………….. II
44 = 4840 / n
n = 4840 / 44
n = 110
Q. Two lamps, one rated 100 W at 220 V and other 60 W at
220 V are connected in parallel to main supply. What current is drawn from the
line if the voltage is 220 V?
Ans: Given,
Power of Lamp 1 = 100 W
Power of Lamp 2 = 60 W
Voltage = 220 V
Resistance of Lamp 1, R1 = V^2/P
= 220^2/
100
= 484
Ω
Resistance of Lamp 2, R2 = V^2/P
= 220^2
/ 60
= 806
Ω
When these two bulbs are connected in Parallel, the
resistance will be equal to:
1/R = 1/R1
+ 1/R2
1/R = R2 + R1 /
R1R2
R = R1R2 / R2 + R1
R = 484 x 806 /
806 + 484
R = 302
Therefore, Current drawn from the Line, I = V / R
= 220
/ 302
=
0.728 A
Q. Which uses more energy, a 250 W TV set for 1 hour or a
1200 W toaster for 10 minutes?
Power
= 250 W
Time
= 1 h i.e; 3600 s
Energy Consumed i.e; Work done = P x T
=
250 x 3600
= 900000
Joules
For Toaster:
Power
= 1200 W
Time
= 10 min i.e; 600 s
Energy Consumed i.e;
Work done = P x T
=
1200 x 600
=
720000 Joules
Therefore, TV uses more energy than Toaster in above case.
Q. An electric heater of 8 Ω resistance draws 15 A from
service mains for 2 hours. Calculate the rate at which heat is developed in the
heater.
Ans: Given;
Resistance
= 8 Ω
Current
= 15 A
Time
= 2 h
Rate at which the heat is developed/ work is done is
equal to Power
We Know,
Power
= V I
P = I
R I (V = IR)
P =
15 x 8 x 15
P = 1800 W
Q. 100J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the Resistor.
Ans: Given:
H = 100 J
R = 4 Ω
T = 1 S
We Know;
H = 12 R T
100 = I2 x 4 x 1
I = √25
I = 5A
Now,
V = I R
V = 5 x 4
V = 20 V
Q. The potential difference between the terminals of an
electric heater is 60V when it draws a current of 4 A from a source. What
current will the heater draw if the potential difference is increased to 120 V?
Ans: Given,
Potential difference = 60 V
Current = 4 A
Thus, Resistance(R)
= V/I
R = 60/4
R = 15 Ohm.
Therefore resistance of heater is 15 Ohm.
Now, When Potential difference is increased to 120 V, the
current will also increase but the resistance will remain same.
Thus New Current (I) = V/R
I = 120/15
I = 8 A
Therefore, 8A current will flow when Potential difference
is increased to 120V.
Q. An electric heater of resistance 20 ohm takes current
of 5 A. Calculate the heat developed in 30 seconds.
Ans: Given,
Resistance (R ) = 20
Current (I) = 5 A
Time = 30 s
We Know,
Heat developed (H) = I^2 x R x T
H = 5 x 5 x 20 x 30
H = 15000 Joules.
Thus, heat developed in the circuit will be 15000 joules.
Q. Let the resistance of an electric component remains
constant while the potential difference across the two ends decreases to half.
What changes will occur in the current through it?
Ans: When the resistance remains constant while the
potential difference decreases to half, the current also reduces to half.
Objective Type Questions
Ans: If the resistance is doubled then current in the circuit become half.
Ans: Copper is used in electric fitting because it is good conductive, highly ductile and inexpensive.
Q. Give the quantitative definition of electric current.
Ans: Quantitatively, Electic current is the rate of flow of electric charge (Q) through a conductor per unit time (t).
Current = Q/t
A. 1/25 B. 1/5 C. 5 D. 25
Ans: D. 25
Q. Which of following terms does not represent electric power in circuit?
A. I2R B. IR2 C. VI D. V2/R
Ans: IR2
A. 100 w b. 75 w c. 50 w d. 25 w
Ans: 25 w
Q. Two conducting wires of same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. The ratio of heat produced in series and parallel combination would be
A. 1:2 b. 2:1 c. 1:4 d. 4:1
Ans: C. 1:4
Ans: In parallel
Q. What is SI unit of conductance
Ans: Seimen
Ans: Alloy of tin and lead
Ans: Tungesten
Ans: Nichrome
Q. In electric fittings we mostly use copper wires? Why
Ans: Because copper is good conductor of electricity.
Q. At the time of short circuit, what happens to the current?
Ans: At the time of short circuit, the current increases multifold. It may cause fire.
Q. The heating elements of electrical heating devices are usually made of:
a. Tungsten b. Bronze c. Nichrome d. Argon
Ans: c. Nichrome
Q. Which of the following characteristics is not suitable for a fuse wire?
a. Thin and Short b. Thick and short c. Low melting point d. Higher resistance
Ans: b. Thick and Short
Q. The heat produced by passing an electric current through fixed resistor is proportional to the square of :
a. Resistance b. Current c. Temperature d. Time
Ans: b. Current
Q. If the current flowing through a fixed resistor is halved, the heat produced in it will become:
a. Double b. One half c. One fourth d. Four times
Ans: c. One fourth
Q. An electric fuse works on the:
A. Chemical effects of current B. Magnetic effects of current
C. Lighting effect of current D. Heating effect of current
Ans: D. Heating effect of Current
Q. Two resistors of resistance 2 ohm and 4 ohm are to be connected to a battery of e.m.f 6V so as to obtain maximum current flowing. Find the current in this case.
Ans: The maximum current will flow in the circuit when there is minimum resistance to the flow of current.
Therefore, the resistances shall be connected in parallel.
Here, V = 6 V,
1/R = ½ + ¼
1/R = 2 + 1
4
1/R = ¾
R = 4/3
R = 1.33
Current (I) = V / R
I = 6/1.33
I = 4.5 A
Q. Find the energy transferred by a 100 W electric bulb in 1 minute.
Ans: We have,
Energy = Power x Time
Energy = 100 w x 60 s
Energy = 6000 Joules
Q. Of which substance is the fuse wire made?
Ans: Fuse wires are made of Nichrome.
Q. In household circuits, is a fuse wire connected in series or in parallel?
Ans: The fuse wire is connected in series in a household circuit.
Q. There is a conductor having the resistivity P. If the conductor is stretched to double of its length, what will be its new resistivity?
Ans: Resistivity does not depend on the length of the conductor. Hence, doubling the length will not change its resistivity.
Q. Electrical resistivity of a given metallic wire depends
upon:
a) its length b) its thickness c) its shape d) nature of material
Ans: d) Nature of material
Q. The device used to measure electric current is called
a) Generator b) Ammeter c) Galvanometer d) Motor
Ans: b) Ammeter
Q. A wire of resistance 1 ohm is divided into two halves
and both halves are connected in parallel. The new resistance will be:
a. 1 ohm b. 2
ohm c. 0.5 ohm d. 0.25 ohm
Ans: d. 0.25 ohm
Q. The energy given to each coulomb of charge passing through a 6V battery is :
a. 12J b. 6J c. 3J d. 18J
Ans: b. 6J
Q. Three resistors having resistances of 2 ohm, 4 ohm and 8 ohm are connected in Parallel. Their equivalent resistance will be:
a. 10 ohm b. 14 ohm c. 8/7 ohm d. Between 2 ohm and 8 ohm
Ans: c. 8/7 ohm
Q. For flow of charges in a conducting metallic wire, the
electrons move only if there is a difference of electric pressure, called the:
(A) Potential difference
(B) Resistance (C) Power (D) Both (A) and (B)
Ans: (A) Potential difference
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