### Electricity

**Question 1**

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)

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)

(b)

(b)

(c) 5

(d) 25

(c) 5

(d) 25

Answer:

(d) 25

Answer:

(d) 25

**Question 2**

Which of the following terms does not represent electrical power in a circuit?

(a) I

(b) IR

(c) VI

(d)

Which of the following terms does not represent electrical power in a circuit?

(a) I

^{2}R(b) IR

^{2}(c) VI

(d)

Answer:

(fa) IR2

Answer:

(fa) IR2

**Question 3**

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be :

(a) 100 W

(b) 75 W

(c) 50 W

(d) 25 W

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be :

(a) 100 W

(b) 75 W

(c) 50 W

(d) 25 W

Answer:

(d) 25 W

Answer:

(d) 25 W

**Question 4**

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. The ratio of heat produced in series and parallel combinations would be :

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

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. The ratio of heat produced in series and parallel combinations would be :

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

Answer:

(c) 1 : 4

Answer:

(c) 1 : 4

**Question 5**

How is a voltmeter connected in the circuit to measure the potential difference between two points ?

How is a voltmeter connected in the circuit to measure the potential difference between two points ?

Answer:

A voltmeter is connected in parallel to measure the potential difference between two points.

Answer:

A voltmeter is connected in parallel to measure the potential difference between two points.

**Question 6**

A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10

A copper wire has diameter 0.5 mm and resistivity of 1.6 x 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 ?

Answer:

If a wire of diameter doubled to it is taken, then area of cross-section becomes four times.

New resistance =
= 2.5 Ω, Thus the new resistance will be
times.

Decrease in resistance = (10 – 2.5) Ω = 7.5 Ω

Answer:

If a wire of diameter doubled to it is taken, then area of cross-section becomes four times.

New resistance =

Decrease in resistance = (10 – 2.5) Ω = 7.5 Ω

**Question 7**

The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below :

The values 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.

Plot a graph between V and I and calculate the resistance of the resistor.

Solution:

The graph between V and I for the above data is given below.

The slope of the graph will give the value of resistance.

Let us consider two points P and Q on the graph.

and from P along Y-axis, which meet at point R.

Now, QR = 10.2V – 34V = 6.8V

And PR = 3 – 1 = 2 ampere

Thus, resistance, R = 3.4 Ω

Solution:

The graph between V and I for the above data is given below.

The slope of the graph will give the value of resistance.

Let us consider two points P and Q on the graph.

and from P along Y-axis, which meet at point R.

Now, QR = 10.2V – 34V = 6.8V

And PR = 3 – 1 = 2 ampere

Thus, resistance, R = 3.4 Ω

**Question 8**

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.

Solution:

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.

Solution:

Here, V = 12 V and I = 2.5 mA = 2.5 x 10

Here, V = 12 V and I = 2.5 mA = 2.5 x 10

^{-3}A**∴**

**Resistance, R =**
=
= 4,800 Ω = 4.8 x 10

^{-3}Ω**Question 9**

A battery of 9V is connected in series with resistors of 0.2 O, 0.3 O, 0.4 Q, 0.5 Q and 12 £1, respectively. How much current would flow through the 12 Q resistor?

A battery of 9V is connected in series with resistors of 0.2 O, 0.3 O, 0.4 Q, 0.5 Q and 12 £1, respectively. How much current would flow through the 12 Q resistor?

Solution:

Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω

Potential difference, V = 9 V

Current through the series circuit, I =
=
= 0.67 A

Solution:

Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω

Potential difference, V = 9 V

Current through the series circuit, I =

**∵**

**There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.**

**Question 10**

How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? [CBSE (Delhi) 2013]

How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? [CBSE (Delhi) 2013]

Solution:

Suppose n resistors of 176 Ω are connected in parallel.

Thus 4 resistors are needed to be connect.

Solution:

Suppose n resistors of 176 Ω are connected in parallel.

Thus 4 resistors are needed to be connect.

**Question 11**

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4Ω

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4Ω

Solution:

Here, R

Solution:

Here, R

_{1}= R_{2}= R_{3}= 6 Ω.**(i) When we connect R**

The equivalent resistance is

_{1}in series with the parallel combination of R_{2}and R_{3}as shown in Fig. (a).The equivalent resistance is

**(ii) When we connect a series combination of R**

_{1}and R_{2}in parallel with R_{3}, as shown in Fig. (b), the equivalent resistance is

**Question 12**

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 ?

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 ?

Solution:

Here, current, I = 5 A, voltage, V = 220 V

Solution:

Here, current, I = 5 A, voltage, V = 220 V

**∴**

**Maxium power, P = I x V = 5 x 220 = 1100W**

Required no. of lamps

Required no. of lamps

**∴**

**110 lamps can be connected in parallel.**

**Question 13**

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 ?

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 ?

Solution:

(i) When the two coils A and B are used separately. R = 24 Ω, V = 220 V

Solution:

(i) When the two coils A and B are used separately. R = 24 Ω, V = 220 V

**(ii) When the two coils are connected in series,**

**(iii) When the two coils are connected in parallel.**

**Question 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.

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.

Solution:

(i) The circuit diagram is shown in figure.

Total resistance, R = 1Ω + 2Ω = 3Ω

Potential difference, V = 6 V

Power used in 2Ω resistor = I

Solution:

(i) The circuit diagram is shown in figure.

Total resistance, R = 1Ω + 2Ω = 3Ω

Potential difference, V = 6 V

Power used in 2Ω resistor = I

^{2}R = (2)^{2}x 2 = 8 W**(ii) The circuit diagram for this case is shown :**

Power used in 2 resistor =
=
= 8 W.

[

Power used in 2 resistor =

[

**∵**

**Current is different for different resistors in parallel combination.]**

**Question 15**

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 ? [CBSE 2018]

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 ? [CBSE 2018]

Solution:

Power of first lamp (P

Potential difference (V) = 220 V

Solution:

Power of first lamp (P

_{1}) = 100 WPotential difference (V) = 220 V

**Question 16**

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

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

Solution:

Energy used by 250 W TV set in 1 hour = 250 W x 1 h = 250 Wh

Energy used by 1200 W toaster in 10 minutes = 1200 W x 10 min

= 1200 x
= 200 Wh 60

Thus, the TV set uses more energy than the toaster.

Solution:

Energy used by 250 W TV set in 1 hour = 250 W x 1 h = 250 Wh

Energy used by 1200 W toaster in 10 minutes = 1200 W x 10 min

= 1200 x

Thus, the TV set uses more energy than the toaster.

**Question 17**

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.

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.

Solution:

Here, R = 8 Ω, 1 = 15 A, t = 2 h

The rate at which heat is developed in the heater is equal to the power.

Therefore, P = I

Solution:

Here, R = 8 Ω, 1 = 15 A, t = 2 h

The rate at which heat is developed in the heater is equal to the power.

Therefore, P = I

^{2}R = (15)^{2}x 8 = 1800 Js^{-1}**Question 18**

Explain the following:

(i) Why is tungsten used almost exclusively for filament of electric lamps ?

(ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal ?

(in) Why is the series arrangement not used for domestic circuits ?

(iv) How does the resistance of a wire vary with its area of cross-section ?

(v) Why are copper and aluminium wires usually employed for electricity transmission?

Explain the following:

(i) Why is tungsten used almost exclusively for filament of electric lamps ?

(ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal ?

(in) Why is the series arrangement not used for domestic circuits ?

(iv) How does the resistance of a wire vary with its area of cross-section ?

(v) Why are copper and aluminium wires usually employed for electricity transmission?

Answer:

(i) The tungsten is used almost exclusively for filament of electric lamps because it has a very high melting point (3300°C). On passing electricity through tungsten filament, its temperature reaches to 2700°C and it gives heat and light energy without being melted.

(ii) The conductors of electric heating devices such as bread-toasters and electric irons, are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that of pure metal and an alloy does not undergo oxidation (or burn) easily even at high temperature.

(iii) The series arrangement is not used for domestic circuits because in series circuit, if one electrical appliance stops working due to some defect, than all other appliances also stop working because the whole circuit is broken.

(iv) The resistance of a wire is inversely proportional to its area of cross-section, i.e., Resistance R

Answer:

(i) The tungsten is used almost exclusively for filament of electric lamps because it has a very high melting point (3300°C). On passing electricity through tungsten filament, its temperature reaches to 2700°C and it gives heat and light energy without being melted.

(ii) The conductors of electric heating devices such as bread-toasters and electric irons, are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that of pure metal and an alloy does not undergo oxidation (or burn) easily even at high temperature.

(iii) The series arrangement is not used for domestic circuits because in series circuit, if one electrical appliance stops working due to some defect, than all other appliances also stop working because the whole circuit is broken.

(iv) The resistance of a wire is inversely proportional to its area of cross-section, i.e., Resistance R

**∝**

**(1/πr**

(v) Copper and aluminium wires usually employed for electricity transmission because they have very low resistances. So, they do not become too hot on passing electric current.

^{2}). If the area of cross section of a conductor of fixed length is increased, then resistance decreases because there are more free electrons for movement in conductor.(v) Copper and aluminium wires usually employed for electricity transmission because they have very low resistances. So, they do not become too hot on passing electric current.

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