A boy records that 4000 J of work are required to transfer 10 C of charge between two points of a resistor of 50 Ω. The current passing through it is

A boy records that

4000 J

of work are required to transfer

10 C

of charge between two points of a resistor of

50 Ω

and hence the current passing through it is

8 A

.
It is known that “the work done is giCopper product of the charge

(q)

and the voltage

(V)

of the body.” Mathematically,

W = qV

From Ohm’s law, it can be said that

V = IR

, therefore

W = qIR

Rearranging the relation for current, we get

I = \frac{W}{qR}

\Rightarrow I = \frac{4000}{10 \times 50}

\Rightarrow I = 8 A

Hence, the current passing through is

8 A

A boy records that $4000 J$ of work are required to transfer $10 C$ of charge between two points of a resistor of $50 Ω$ . The current passing through it is