# Current Electricity Class 12 Numericals

## Q.1. What will be time taken of current drawn of 0.741 micro-ampere by the net value if electric charge is 100 Coulomb?

**Ans: **we know that the Current is defined as time rate of flow of charge.

i.e. I = q/ t ...............................(1)

**Given in the question : **

I = 0.741 microampere = 0.742 × 10^-6 A

q = 100 C

From (1),

t = q/ I

= 100/ 0.741 × 10^-6 s

= 134.95 × 10^6 s

= **1.35 × 10^8 s**

## Q.2 . An electric bulb is drawing a current of 0.5 A from a 220 V generator. Find the power of the bulb ?

**Ans : **

**Given in the question :**

Voltage, V = 220 V

Current, I = 0.5 A

We know that the power consumed by an electric device is given by

P = VI

= 220 × 0.5

= **110 W**

## Q.3. Find the current drawn by the combination of 4 resistors through the battery as shown :

**Ans :**In the above network, the resistor between AB is in parallel with the series combination of other three resistors.

Hence, the equivalent resistance of three resistors between AD, DC and CB is

Re = 10 + 10 + 10

= 30 Î©

Now, this resistance R is in parallel with the resistance between AB . Therefore the resultant resistance is given by

R = 1/30 + 1/10

= 30 / (1 + 3)

= 30 / 4

= 7.5

**Î©**

Hence, the current drawn from the battery is

I = V /R

= 3 / 7.5

= 30 / 75

=

**0.4 A**

## Q.4. If a current of 1 A is flowing through a wire for 1 S, find the number of electrons passing through any cross section of the wire.

**Ans :**

__Given in the question :__

Current, I = 1 A

Time, t = 1 s

We know that current flowing through a wire is given by

I = q/ t

or, q = It

or, = 1× 1 = 1 C

**Since there are 6.28 × 10^18 electrons in 1 C, hence 6.28 × 10^18 electrons are passing through the wire.**

## Q.5. An electron is released from rest in uniform electric field of 10^6 N/C. Calculate its acceleration. Also find the time taken by the electron in attaining a speed of 0.1 C where is 3× 10^8 m/s.

**Ans :**

__Given in the question :__

Electric field , E = 10^6 N/C

Velocity, v = 0.1 × 3 × 10^ 8 m/s

We know that the acceleration of an electron in an electric field is given by

a = eE/ m

= (1.6 × 10^-19 × 10^6)/ 9.1 × 10^-31

= 0.175 × 10^18

= 1.75 × 10^17 m/ s^2

Again we have,

v = u +at

Since initial velocity u is zero ( u = 0 )

Therefore, v = at

or, t = v / a

or, t = 3 × 10^7 / 1.75 × 10^17

or,

**t = 1.71 × 10^-10 s**

**Get notes of Current Electricity class 12 : Current Electricity**

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