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