# Maharashtra Board Class 11 Physics Chapter 11 Electric Current Through Conductors Solution

## Maharashtra Board Class 11 Physics Solution Chapter 11 – Electric Current Through Conductors

Balbharati Maharashtra Board Class 11 Physics Solution Chapter 11: Electric Current Through Conductors. Marathi or English Medium Students of Class 11 get here Electric Current Through Conductors full Exercise Solution.

 Std Maharashtra Class 11 Subject Physics Chapter 11 Chapter Name Electric Current Through Conductors

1.) Choose correct alternative

i) You are given four bulbs of 25 W, 40 W, 60 W and 100 W of power, all operating at 230 V. Which of them has the lowest resistance?

(A) 25 W (C) 40 W

(C) 60 W (D) 100 W

We have

Power = v2/R

P ∝ 1/R

Lowest resistance is corresponding to highest power hence the 100W has lowest resistance.

ii) Which of the following is an ohmic conductor?

(A) transistor(B)vacuum tube

(C) electrolyte(D) nichrome wire

All other except nichrome wire are bad conductor.

iii) A rheostat is used

(A) to bring on a known change of resistance in the circuit to alter the current

(B) to continuously change the resistance in any arbitrary manner and there by alter the current

(C) to make and break the circuit at any instant

(D) Neither to alter the resistance nor the current

Rheostat is variable resistance which changes continuously.

iv) The wire of length L and resistance R is stretched so that its radius of cross-section is halved. What is its new resistance?

(A) 5R (B) 8R

(C) 4R (D) 16R

v) Masses of three pieces of wires made of the same metal are in the ratio 1:3:5 and their lengths are in the ratio 5:3:1. The ratios of their resistances are

(A) 1:3:5 (B) 5:3:1

(C) 1:15:125 (D) 125:15:1

We know that

vi) The internal resistance of a cell of emf 2V is 0.1Ω it is connected to a resistance of 0.9Ω. The voltage across the cell will be

(A) 0.5 V (B) 1.8 V

(C) 1.95 V (D) 3V

We have formula for emf

E = I(R+r)

I = E/(R+r)

I = 2/(0.9+0.1)

I = 2/1

I= 2 A

VOLTAGE IS GIVEN BY

V= IR

V= 2*0.9=1.8 V

vii) 100 cells each of emf 5V and internal resistance 1Ω are to be arranged so as to produce maximum current in a 25Ω resistance. Each row contains equalnumber of cells. The number of rows should be

(A) 2 (B) 4

(C) 5 (D) 100

viii) Five dry cells each of voltage 1.5 V are connected as shown in diagram

What is the overall voltage with this arrangement?

(A) 0V (B) 4.5V

(C) 6.0V (D) 7.5V

Overall voltage= voltage in same direction- voltage in opposite direction.

Overall voltage= 4´1.5-1.5

Overall voltage=3´1.5

= 4.5V

i) In given circuit diagram two resistors are connected to a 5V supply.

a] Calculate potential difference across the 8Ω resistor.

b] A third resistor is now connected in parallel with 6Ω resistor. Will the potential difference across the 8Ω resistor the larger, smaller or the same as before? Explain the reason for your answer.

a] Here resistances are in series therefore Rs= 8+6=14Ω

total current flowing is given by

I = V/Rs = 5/14 = 0.36 A

Therefore potential difference across the 8Ω resistor is given by

V = IR = 0.36 ´ 8 = 2.88 V

b] A third resistor is now connected in parallel with 6Ω resistor. Then diagram is given

We know that parallel resistance in circuit decreases the overall resistance hereby increasing overall current through circuit. Hence voltage across 8 ohm will also increase due to increase in overall current.

ii) Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.

Consider a metallic conductor having length L.

Total charge on metallic conductor is given by,

Total charge = number of free electrons ´ charge on one electron ………1

Let n be the number of free electrons per unit volume then

n = (number of free electrons)/volume

Number of free electrons = n X V

Electrons = n X A X L