Question

2. An electric appliance does work of 250 J when charge
of 7 C moves across the two points of the conductor
Calculate the voltage applied.​

Answer 1

Answer:

$\large{\underline{\boxed{\sf 35.71 \: V}}}$

Explanation:

Given that ;

Work done by the electrical appliances is 250 joule.

⇒ Work Done = 250 J

Charge (Q) = 7 coulomb

To find: The voltage applied.

Solution:

We know that ;

Work done is given by the product of voltage applied and charge.

⇒ $\mathrm{W = V * Q}$

⇒ $\mathrm{250 = V * 7}$

⇒ $\text{V} = \dfrac{250}{7}$

⇒ $\text V = 35.71$

Hence, the voltage applied is 35.71V.

Answer 2

HeRe Is Your Ans ⤵

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Answer :-

➡$\fbox{ \fbox{ \: V = 35.714 \: volt \: }}$

Given :-

➡Work Done ( W ) = 250 Joule

➡Charge ( Q )= 7 Coulomb

To Find :-

➡Potential Difference ( V ) = ?

Formula :-

➡V = W ÷ Q

Solution :-

➡V = W ÷ Q

➡V = 250 ÷ 7

➡$\fbox{ \fbox{ \: V = 35.714 \: volt \: }}$

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