- Introduction
- |
- Electric Charges
- |
- Conductors and insulators
- |
- Electric potential and potential difference
- |
- Electric current and electrical circuits
- |
- Circuit Diagrams
- |
- Ohm's Law
- |
- Factors affecting of resistances of a conductor
- |
- Resistance of a system of resistors
- |
- Heating Effect of current
- |
- Applications of heating effect of current
- |
- Electric Power

- Electric Charge
- |
- Electric Potential
- |
- Materials(conductors, insulators & superconductors)
- |
- Electric Current
- |
- Ohm's Law

In this page we have *Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions* . Hope you like them and do not forget to like , social share
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(a) 1/25

(b) 1/5

(c) 5

(d) 25

Let’s now justify this answer

Resistance of a piece of wire is proportional to its length. According to question we have a piece of wire of resistance R. The wire is cut into five equal parts. Therefore resistance of each part would be

Since

All five resistance are connected in parallel. So, for parallel resistance

Therefore the ratio of R/R’ is 25.

(a) I

(b) IR

(c) VI

(d) V

Electric power is given by the expression

(i)

According to Ohm’s Law

(ii)

Where,

V = Potential difference

I = Current

R = resistance

Equation (i) can be written as

Again considering equation (ii)

Power can not be expressed as IR

Energy consumed by an appliance is given by th expression

Where,

Power rating, P=100 W

Voltage, V=220 V

Resistance,

The resistance of the bulb remains constant if the supply voltage is reduced to 110 V.

If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as

Therefore , the power consumed would be 25 W.

(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1

So, for series combination

Power in series combination is

For parallel combination of resistances

Power in parallel combination is

So,

D=0.5 mm,

Using

If the diameter is doubled radius of copper wire is also doubled.

For the same length and same material wire,

We get

Hence, the resistance becomes one fourth of the original one.

I (amperes) |
0.5 |
1.0 |
2.0 |
3.0 |
4.0 |

V (volts) |
1.6 |
3.4 |
6.7 |
10.2 |
13.2 |

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

The plot between voltage and current is called VI characteristic. The voltage is plotted on x-axis and current is plotted on y-axis.

Resistance (

Where,

Potential difference,

Current in the circuit,

Therefore, the resistance of the resistor is 4.8 kΩ

We know that when the resistors are connected in series, same current through all the resistor and there is no current divisioning happening.

Current flow through the component is the same, given by Ohm’s law as

Where,

Now These are connected in series. Hence, the sum of the resistances will give the value of

Potential difference,

Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.

For

Where,

Supply voltage,

Current,

Equivalent resistance of the combination =

1/R

1/R

R

Now from Ohm’s Law

Therefore, four resistors of 176 Ω are required to draw the given amount of current.

This is quite interesting question; we need to look for the arrangement where desired result can be obtained.

Let’s check out few arrangement

a) If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω

This is not what we desire

b) If we connect the resistors in parallel, then the equivalent resistance will be 6/2 = 3 Ω

This is not what we desire

c) Two resistor in parallel

Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be

The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω+ 3 Ω = 9 Ω.

This is what we desire in part (i)

(b) Two resistor in series

Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω.

The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be

Therefore, the total resistance is 4 Ω

This is what we desire in part (ii)

aximum allowable current, I = 5 A

Rating of an electric bulb P=10 watts

Total resistance of the circuit is

Now resistance of each bulb

Let n be the number of bulbs in the circuit in parallel, then

Supply voltage, V= 220 V

Resistance of one coil, R= 24

(i) Coils are used separately

According to Ohm's law,

V= I

Where,

I

Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series

Total resistance for coils connected in series is given by expression

Total resistance of coils connected in series is

According to Ohm's law,V = I

Where, I

Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.

(iii) Coils are connected in parallel and equivalent resistance of parallel combination is given as

Total resistance, R

According to Ohm's law,

V=I

Where,

I

Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.

Answer.

(i) It is given in the question that,

Potential difference, V = 6 V

And 1Ω and 2Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1Ω + 2Ω = 3Ω

According to Ohm's law,

V = IR

Where,

I is the current through the circuit

This current will flow through each component of the circuit because current is not divided between resistors in series combination of circuits. Hence, current flowing through the 2 resistor is 2 A. Power is given by the expression,

P= (I)

(ii) Again from second part of the question it is given that,

Potential difference, V = 4 V

And 12Ω and 2Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2Ω resistor will be 4 V.

Power consumed by 2Ω resistor is given by

Comparison between the power used in both the cases is

Therefore, potential difference across each of them will be 220 V,

This happens because no division of voltage takes place in a parallel circuit.

Current drawn by the bulb of rating 100 W is given by,

Power = Voltage x Current

Hence, current drawn from the line would be,

Where, P = power of the appliance and t = time.

Energy consumed by a TV set of power 250W in 1 hour = 250 W ? 3600 s = 9?10

Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600= 7.2×10

Therefore , the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.

Where resistance of th electric heater is

Current drawn is I=15A

Power P is,

Therefore heat is produced by the heater is at the rate of 1800J/s.

(a) Why is the tungsten 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?

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

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

(e) Why are copper and aluminum wires usually employed for electricity transmission?

(a) The melting point and resistivity of tungsten are very high. It does not burn readily at a high temperature. The electric lamps glow at very high temperature. This is the reason behind using tungsten as heating element of electric bulbs.

(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloys because resistivity of alloys are more then that of metals. It produces large amount of heat.

(c) In series arrangement of resistances if the circuit breaks at one point then the flow of current would stop in whole circuit. This is the reason why we do not use series circuit in domestic circuits.

(d) Resistance of a wire is inversely proportional to its area of cross-section, that is when area of cross-section of wire increases , the resistance of the wire decreases and vice-versa.

(e) Copper and aluminum wires have low resistivity. They are good conductors of electricity. Hence they are u

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