Question

I want 5 extra numerical in the lesson electricity class 10 physics

Answer 1

Given below are the Class 10 Solved Numerical Questions for electricity

Question 1

A wire of length 3 m and area of cross-section 1.7 × 10-6 m2 has a resistance 3 × 10-2 ohm.

a. What is the formula for resistivity of the wire and what is the unit of it

b. Calculate the resistivity of the wire

Solution

a. Resistivity of the wire is given by

ρ=R

A

L

And It unit is Ohm-m

b. In this case

L=3 m

A=1.7×10−6 m2

R=3×10−2 ohm

So

ρ=1.7×10−8 Ohm-m

Question 2

The table given below shows the resistivity of three Material X, Y and Z?

Samples

X

Y

Z

Resistivity

3 X 10-9

11.1 X 0-6

18X10-17

a. Arrange the samples in increasing order of conductivity

b. Which of these is best conductor?

c. Which are these is best insulator?

Solution

a.Conductivity is inversely proportional to resistivity

So

Y< X< Z

b. Z is the best conductor as it has least resistivity

c. Y is the best insulator as it has highest resistivity

Question 3.

There are m resistor each of resistance R. First they all are connected in series and equivalent resistance is X. Now they are connected in parallel and equivalent resistance is Y. What is the ratio of X and Y?

Solution

Series combination

X=R+R+R+......mtimes=mR

Parallel combination

1

Y

=

1

R

+

1

R

+

1

R

+...=

m

R

Or Y=

R

m

So X : Y=m2 : 1

Question 4

We have four resistors A ,B ,C and D of resistance 4 ohm,8 ohm ,12 ohm and 24 ohm respectively?

1

Lowest resistance which can be obtained by combining these four resistors

2

highest resistance which can be obtained by combining these four resistors

Solution

Lowest resistance is obtained in parallel combination

1

R

=

1

4

+

1

8

+

1

12

+

1

24

Or R=2Ω

Highest resistance is obtained in series combination

R

=

4

+

8

+

12

+

24

=

48

Ω

Question 5

Three resistors 5Ω, 10Ω and 30Ω are connected in parallel with the battery of Voltage 6V?

S.no

Questions

1

The value of current across each resistor

2

The value of Potential difference across each resistor

3

Total current in the circuit

4

Effective resistance of the circuit

Solution

Potential difference remains same across parallel combination

So current in each resistor is calculated as

I1=V/R1=6/5=1.2 A

I2=V/R2=6/10=.6 A

I3=V/R3=6/30=.2 A

Total current in the circuit

I=I1+I2+I3=1.2+.6+.2=2 A

Effective resistance

1

R

=

1

5

+

1

10

+

1

30

R= 3Ω