- 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
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- 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 *NCERT solutions for class 10 Science: Chapter 12 electricty Intext Questions* . Hope you like them and do not forget to like , social share
and comment at the end of the page.

1 A = 1Cs

q = ne

where n is any integer positive or negative

If q=1C and e= 1.6 ×10

Then,

Or,

So,

This way we can calculate the number of electrons on charged body if we know the total charge on the charged body.

The expression to be used in solving this question is

Here according to question, Charge = 1C and Potential difference=6V

So, Work done = 6×1= 6J

Therefore, 6J of energy is given to each Coulomb of charge passing through a battery of 6 V.

- Material of the conductor
- Temperature of the conductor
- Length of the conductor
- Area of cross-section of the conductor

From above equation we can see that resistance is inversely proportional to the area of cross-section of the wire. Thick wire means more area of cross-section and lower the resistance of wire. Similarly thin wire means less area of cross-section and wire would have higher resistance.

Therefore, current can flow more easily through thick wire than a thin wire.

Now according to question the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. So, we now have

Therefore from Ohm’s law we can obtain the

Therefore the amount of current flowing through the electric component is reduced to half.

(a) resistivity of an alloy is generally higher then that of constituent metals.

(b) alloys have high melting point.

(c) alloys do not oxidize.

to answer the following –

(a) Which among iron and mercury is a better conductor?

(b) Which material is the best conductor?

Resistivity of mercury = 94.0 x 10

Comparing the resistivity of both iron and mercury we conclude that , resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.

(b) It can be observed from above Table that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.

Three cells of potential 2 V, each one of these cells are connected in series therefore the potential difference of the combined battery will be

2 V + 2 V + 2 V = 6V.

The following circuit diagram shows three resistors of resistances 5Ω , 8Ω and 12Ω respectively connected in series and a battery of potential 6 V and a plug key which is closed means the current is flowing in the circuit.

difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?

The resistances are connected in series.Ohm's law can be used to obtain the readings of ammeter and voltmeter. According to Ohm's law,

V = IR,

Where, Potential difference, V = 6 V

Current flowing through the circuit/resistors = I

Resistance of the circuit, R = 5Ω + 8Ω + 12Ω = 25Ω

I = V/R = 6/25 = 0.24 A

Potential difference across 12Ω resistor = V

Current flowing through the 12Ω resistor, I = 0.24A

Therefore, using Ohm's law, we obtain

V

Therefore, the reading of the ammeter will be 0.24 A.

The reading of the voltmeter will be 2.88 V.

We know that for parallel combination of resistances equivalent resistance is given by the relation

If R is the equivalent resistance of the parallel combination then

Or,

Therefore, equivalent resistance is nearly equal to 1Ω.

(b) When 1Ω, 10

Let R be the equivalent resistance

0r,

Which is the equivalent resistance of given resistances.

Hence, in both given cases , the equivalent resistance is less than 1Ω

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?

Resistance of toaster, R

Resistance of water filter, R

Potential difference of the source, V = 220V

It is given in the question that these are connected in parallel combination. Equivalent resistance of resistances in parallel combination is

Let R be the equivalent resistance of the circuit.

According to Ohm’s Law , V=IR

Or,

Where I is the current flowing through the circuit. So, we have

Therefore the current drawn by an electric iron connected to the same source of potential 220 V = 7.04 A

Let R’ be the resistance of the electric iron. According to Ohm’s law,

Therefore the resistance of the electric iron is 31.25Ω and the current flowing through it is 7.04 A.

them in series?

- There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.
- The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.
- If one appliance fail to work, other will continue to work properly if they are connected in parallel combination.

(a) In order to get 4Ω , resistance 2Ω resistance should be connected in series with the parallel combination of 3Ω and 6Ω resistances as shown below in the figure.

This gives

(b) In order to get 1Ω all three resistors should be connected in parallel combination as shown in this figure

Equivalent resistance would be

(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by

Also heat produced ‘H’ is given by the relation

Thus, for same amount of current

From this we can conclude that more heat is produced by the heating element as it has more resistance, this is the reason it glows.

difference of 50 V.

Given Charge, Q = 96000C,

Time, t= 1hr

Potential difference, V= 50 Volts

Now we know that

R=20Ω ; I= 5A and t=30s

Where,

Voltage,V = 220 V

Current, I = 5 A

P= 220 x 5 = 1100 W

Energy consumed by the motor = Pt

Where,

Time, t = 2 h

Energy consumed = VIt = Pt = 1100 x 2 = 2200 Wh

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