Question

What is the work done to move charges in a circuit having 6 V potential difference, in which 10 A current flows for ½ minute?

Answer 1

Answer:

Given:

Potential Difference = 6V

Current = 10 amp

Time = ½ minute

To find:

Work done to move the charge in the circuit

Concept:

First we will try to find out the amount of charge passing through the circuit in the given time.

Then as per the definition of potential Difference, we can calculate the work done.

Potential Difference is defined as the work done to transport unit positive charge from one point to another.

Calculation:

$current = \dfrac{charge}{time}$

$= > 10 = \dfrac{charge}{30}$

$= > charge = 300 \: coulomb$

As per definition of Potential Difference :

$PD = \dfrac{work}{charge}$

$= > 6 = \dfrac{work}{300}$

$= > work = 6 \times 300$

$= > work = 1800 \: joule$

$= > work = 1.8 \: kJ \:$

So final answer :

$\boxed{ \red{ \sf{ \bold{ \huge{work = 1.8 \: kJ \: }}}}}$

Answer 2

Answer:

1800 J

Explanation:

We have given :

Potential difference V = 6 V

Current I = 10 A

Time t = 1 / 2 min or 30 sec

We know :

I = Q / t

Q = I × t

Putting values here we get :

Q = 10 × 30 C

= > Q = 300 C

We are asked to find work done :

We have expression :

Potential difference = Work done / Charge

= > W = V × Q

Putting vales here we get :

= > W = 6 × 300 J

= > 1800 J

Hence we get required answer.