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Electricity : NCERT Intext Questions


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Q.1       What does an electric current mean?
Sol.       
The flow of electric charge is known as electric current. Electric current is carried by moving electrons through a conductor. Electric current flows in opposite direction to the movement of electrons.


Q.2       Define the unit of current.
Sol.       
SI unit of electric current is ampere (A).

Ampere is the flow of electric charges through a surface at the rate of one coulomb per second, i.e. if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.
Therefore, 1 ampere = {{1\,C} \over {1\,s}}


Q.3       Calculate the number of electrons constituting one coulomb of charge.
Sol.        
We know that charge over 1 electron = 1.6 × 10–19 coulomb

Thus, 1.6 × 10–19 C of charge = 1 electron
Therefore, 1 C of charge = {1 \over {1.6 \times {{10}^{ - 19}}}} Electrons
= {{{{10}^{19}}} \over {1.6}} electrons = {{10 \times {{10}^{18}}} \over {1.6}} electrons = 6.25 × 1018 electrons


Q.4       Name a device that helps to maintain a potential difference across a conductor.
Sol.       Battery or a cell


Q.5       What is meant by saying that the potential difference between two points is 1 V?
Sol.       This means 1 joule of work is done to move a charge of 1 coulomb between two points.


Q.6       How much energy is given to each coulomb of charge passing through a 6 V battery?

Sol.        Given, Charge Q = 1C, Potential difference, V = 6V

Therefore, Energy i.e. Work done, W =?
We know that, V = {W \over Q}
Therefore, 6V = {W \over {1C}}
 \Rightarrow W = 6V × 1C = 6J
Thus, required energy = 6J


Q.7       On what factors does the resistance of a conductor depend?
Sol.        
Resistance of a conductor depends upon:

(a) Nature of conductor
(b) Length of conductor
(c) Area of cross section of conductor


Q.8       Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Sol.       
Since, resistance is indirectly proportional to the area of cross section, thus current flows easily through a thick wire compared to a thin wire of the same material.


Q.9       Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Sol.       
Since Resistance (R) = {{Potential\,Difference(V)} \over {Electric\,current\,(I)}}

Therefore, if potential difference between two ends of the component will be halved, and resistance remains constant, then electric current would also be halved.


Q.10     Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Sol.        
Since, alloys have higher melting point than pure metal so coils of electric toasters and electric irons are made of an alloy rather than a pure metal to retain more heat without melting.


Q.11     Use the data in Table 12.2 to answer the following –
             (a) Which among iron and mercury is a better conductor?
             (b) Which material is the best conductor?
Sol.         
(a) Iron

               (b)Silver


Q.12     Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Sol.

 


Q.13     Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Sol.

 

The total resistance in the circuit = Sum of the resistances of all resistors
= 5 Ω + 8 Ω + 12 Ω = 25 Ω
We know;
R={V \over I} or, 25\Omega = {{6V} \over I} or, I = {{6V} \over {25\Omega }} = 0.24A
Since, resistances are connected in series, thus electric current remains the same through all resistors.
Here we have,
Electric current, I = 0.24A
Resistance, R = 12Ω
Thus, potential difference, V through the resistor of 12Ω = I x R
Or, V = 0.24A x 12Ω = 2.88 V
Thus, reading of ammeter = 0.24A
Reading of voltmeter through resistor of 12Ω = 2.88V


Q.14     Judge the equivalent resistance when the following are connected in parallel –
             (a) 1 Ω and 106 Ω,                       (b) 1 Ω and 103 Ω, and 106 Ω.
Sol. 

Since {1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}} + {1 \over {{R_3}}} + .. + {1 \over {{R_n}}}
when resistors are connected in paralle
(a) 1 Ω and 106 Ω
Thus, {1 \over R} = {1 \over {1\Omega }} + {1 \over {106\Omega }} = {{106 + 1} \over {106}}\Omega = {{107} \over {106}}\Omega
Thus, R = {{106} \over {107}}\Omega = 0.99\Omega
Thus, equivalent resistance of 1\Omega and 106\Omega are connected in parallel = 0.99\Omega
(b)  1 Ω and 103 Ω, and 106 Ω
Thus, {1 \over R} = {1 \over {1\Omega }} + {1 \over {103\Omega }} + {1 \over {106\Omega }} = {{10918 + 106 + 103} \over {103 \times 106}}\Omega = {{11127} \over {10918}}\Omega
Thus, R = {{10918} \over {11127}}\Omega = 1.02\Omega
Thus, equivalent resistance of  1 Ω, 103 Ω and 106 Ω are connected in parallel = 1.02\Omega


Q.15     An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source.
What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Sol.

Given:
R1 = 100Ω, R2 = 50 Ω, R3 = 500 Ω all are connected in parallel
Potential difference = 220V
Thus, {1 \over R} = {1 \over {100\Omega }} + {1 \over {50\Omega }} + {1 \over {500\Omega }} = {{5 + 10 + 1} \over {500}}\Omega = {{16} \over {500}}\Omega
Therefore, R = {{500} \over {16}}\Omega = 31.25\Omega
Electric current (I) through the circuit = {V \over R}
 \Rightarrow I = {{220V} \over {31.25\Omega }} = 7.04A
For electric iron
Since it takes as well current as three appliances, thus electric current through it = 7.04A
The electric current = 7.04 A and potential difference = 220 V
Thus, Resistance of electric iron = Total resistance of three appliances = 31.25 Ω
Thus, electric current through the electric iron = 7.04A
Resistance of electric iron = 31.25 Ω


Q.16    What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Sol.

Advantages of connecting electrical appliances in parallel instead of connecting in series:
(a) Voltage remains same in all the appliances.
(b) Lower total effective resistance


Q.17     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?
Sol.

(a) When resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series

Let total resistance due to resistors having resistance equal to 6 Ω and 3 Ω = R1
Therefore, {1 \over {{R_1}}} = {1 \over {6\Omega }} + {1 \over {3\Omega }} = {{1 + 2} \over 6}\Omega = {3 \over 6}\Omega = {1 \over 2}\Omega
Thus {R_1} = 2\Omega
Now, total effective resistance in the circuit = R1 + 2 Ω = 2 Ω + 2 Ω = 4 Ω
Hence, when resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series, then the total effective resistance in the circuit = 4 Ω

(b) When all the three resistance is connected in parallel then
{1 \over R} = {1 \over {2\Omega }} + {1 \over {3\Omega }} + {1 \over {6\Omega }} = {{3 + 2 + 1} \over 6}\Omega = {6 \over 6}\Omega = 1\Omega
Thus, R = 1\Omega
When all the three resistance will be connected in parallel, then the total effective resistance in the circuit = 1 Ω


Q.18     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?
Sol.

When all the resistors are connected in series
Then the {1 \over R} = {1 \over {4\Omega }} + {1 \over {8\Omega }} + {1 \over {12\Omega }} + {1 \over {24\Omega }} = {{6 + 3 + 2 + 1} \over {24}}\Omega = {{12} \over {24}}\Omega = {1 \over 2}\Omega
Thus, R = 2 Ω
(b) When all the resistors are connected in series
Thus, total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω
Thus, highest resistance = 48 Ω
Lowest resistance = 2 Ω
Page Number – 218


Q.19    Why does the cord of an electric heater not glow while the heating element does?
Sol.

The resistance of heating element is very high compared to the cord of an electric heater. Since, heat produced is directly proportional to the resistance, thus element of electric heater glows because of production of more heat and cord does not.


Q.20    Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Sol.

Given, Potential difference (V) = 50V,
Charge (Q) = 96000 coulomb
Time, t = 1 hour = 60 X 60 s = 3600 s
Heat produced =?
We know that electric current (I) = {Q \over t} = {{96000C} \over {3600s}} = {{96000} \over {3600}}A
We know that heat produced (H) in the given time (t) = VIt
H = 50V \times {{96000} \over {3600}}A \times 3600 = 50 \times 96000J = 4800000J = 4.8 \times {10^6} Joule


Q.21     An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Sol.

Given, Electric current (I) = 5A, Resistance (R) = 20 Ω, Time (t) = 30s
Therefore, Heat produced (H) =?
Since, V = I \times R = 5A \times 20\Omega = 100V
We know that, Heat produced (H) = VIt
 \Rightarrow H = 100V \times 5A \times 30s = 15000J = 1.5 \times {10^4}J Answer


Q.22     What determines the rate at which energy is delivered by a current?
Sol.

Since electric power is the rate of consumption of electric energy in any electrical appliance. Hence, rate at which energy is delivered by a current is the power of electric appliance.


Q.23     An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Sol.       
Given, Electric current (I) = 5A,

Potential difference (V) = 220V,
Time (t) = 2h = 2 x 60 x 60 s = 7200 s
Power (P) =?
Energy consumed =?
We know that, P = VI = 220V × 5A = 1100W
We know that energy consumed by the electric appliance = P × t
 \Rightarrow Energy consumed = 1100 W × 7200 s = 7920000 J = 7.92 × 106 J  



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