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Saturday, 27 June 2020

CBSE Class 10th Electricity: NCERT Solution from 11 to 20 Part-2

                             Electricity: NCERT Solution



11.     How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer:Given, resistance of each of the resistor = 176 Ω
Electric current ( I ) = 5A
Potential difference (V) = 220V
Number of resistors connected in parallel =?

Let total x resistors are connected in parallel, and total effective resistance = R


Thus, 4 resistors need to be connected.

 

12.    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. 

Answer: Here we have four options to connect the three resistors in different ways.

·         All the three resistors can be connected in series

·         All the three can be connected in parallel

·         Two of the three resistors can be connected in series and one in parallel and

·         Two of the three resistors can be connected in parallel and one in series

Thus, effective resistance in the case

When all the three resistors are connected in series

Effective total resistance = 6 Ω + 6 Ω + 6 Ω = 18 Ω. This is not required

When all the three are connected in parallel


Thus, effective total resistance R = 2 Ω. This is also not required.

When two of the three resistors are connected in parallel and one in series

When two resistors are connected in parallel


Therefore, R = 3Ω

And third one is connected in series, then total resistance = 3 Ω + 6 Ω = 9 Ω

Two of the three resistors can be connected in parallel and one in series

When two resistors are connected in series, then total resistance = 6 Ω + 6 Ω = 12 Ω

And one resistor is connected in series with two in parallel


Thus, R = 4 Ω

Thus, when two resistors are connected in series and one in parallel then total effective resistance = 9 Ω

When two resistors are connected in parallel with one in series then total effective resistance = 4 Ω

13.    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 with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Answer: Given, potential difference (V) = 220 V
Power input (P) = 10W
Allowable electric current (I) = 5A
Number of lamps connected in parallel =?

To calculate this, first of all resistance of each of the lamp is to be calculated.

We know; P = V2 ÷ R

Hence, 10 W = (220 V)2 ÷ R

Or, R = 48400 ÷ 10 W = 4840 Ω

Let x bulb are to be connected in parallel to have electric current equal to 5 A. Therefore;


Thus, total 110 bulbs need to be connected in parallel.

14.    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, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer: Given, Potential difference (V) = 220 V
Resistance of each coil = 24 Ω

The current in given three case, i.e. when used separately, when used in parallel, when used in series =?

Case 1: When used separately, then resistance, R = 24 Ω and V = 220V

We know; I = V/R

= 220 V ÷ 24 Ω = 9.16 A

Case 2: When the two resistors are connected in series.

Total effective resistance = 24 Ω + 24Ω = 48 Ω

Hence, electric current can be calculated as follows:

I = V/R = 220 V ÷ 58 Ω = 4.58 A

Case 3: When the two resistors are connected in parallel,


Therefore, total effective resistance R = 12 Ω

Thus, electric current I = V/R

= 220 V ÷ 12 Ω = 18.33 A

Thus, electric through the circuit

When coil is used separately = 9.16A

When coils are used in series = 4.58 A

When coils are used in parallel = 18.33 A

 

15.    Compare the power used in the 2 Ω resistor in each of the following circuits:

a.     A 6 V battery in series with 1 Ω and 2 Ω resistors, and 

Answer: Potential difference = 6 V
Resistance of resistors = 1 Ω and 2 Ω 
Power used through resistors of 2 Ω =?

Since, resistors are connected in series, thus, total effective resitance, R = 1 Ω + 2 Ω = 3 Ω

We know; I = V/R = 6 V ÷ 3 Ω = 2 A

Since current remains same when resistors are connected in series, hence current through resistor of 2 Ω = 2 A

Thus, power (P) = I2 x R

= (2 A)2 x 2 Ohm = 8 W

b.    A 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer: Potential difference, V = 4V
Resistance of resistors connected in parallel = 12 Ω and 2 Ω
Power used by resistors having resistance of 2 Ω = ?

Since, voltage across the circuit remains same if resistors are connected in parallel.

Thus, power (P) used by resistance of 2 Ohm

= V2/R = (4 V)2/2 Ω

= 16/2 = 8 W

Thus, power used by resistance of 2 Ω in both the case = 8 W

16.    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? 

Answer: Since, both the lamps are connected in parallel, thus, potential difference will be equal

Thus, potential difference = 220 V
Power of one lamp, P1 = 100W
Power of second lamp, P2 = 60W

We know; P = VI

Or, I = P/V

Thus, total current through the circuit


17.    Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer: Given, power of TV (P) = 250W,
time (t) = 1 hr = 60 x 60 s = 3600 s
Thus, energy used by it =?

We know that energy used by appliance = Power x time

Thus, energy used by TV = 250 W x 3600 s = 900000 J = 9 x 105 J

Power of toaster = 1200W
Time (t) = 10 minute = 60 x 10 = 600 s

Thus, energy used by toaster = P x t = 1200W x 600 s = 720000 J = 7.2 x 105 J

Thus, given TV set will use more energy than toaster.

18.    An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Answer: Given, Resistance, R = 8 Ω
Electric current ( I ) = 15A
Time (t) = 2 h = 2 x 60 x 60 s = 7200 s

Rate at which heat is developed in heater =?

We know that rate of heat produced = I2R = (15A)2 x 8 Ω = 225 x 8 J/s = 1800 J/s Answer

19.    Explain the following.

a.     Why is tungsten used almost exclusively for filament of electric lamps?

Answer: The melting point of tungsten is very high, i.e. 3380°C, which enables it not to melt at high temperature and to retain most of the heat. The heating of tungsten makes it glow. This is the cause that tungsten is used almost exclusively for filament of electric lamps.

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?

Answer: To produce more heat, the high melting point of conductors is necessary. The alloys of metal have higher melting points than pure metals. Thus, to retain more heat alloy is used rather than pure metal in electrical heating devices, such as bread-toaster, electric iron, etc.

c.     Why is the series arrangement not used for domestic circuits? 

Answer: There is loss of voltage in the series arrangement in the circuits because of add on effect of resistances. So, series arrangement is not used for domestic circuits.

20. How does the resistance of a wire vary with its area of cross-section?

Answer: Resistance of a wire is indirectly proportional to the area of cross section. Resistance increases with decrease in area of cross section and vice versa.

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