Question

A battery of 10 volt carries 20000c of charge through a resistance 20 ohm. The work done in 10 seconds is?

Answer 1

★ Given :

Potential Difference (v) = 10v

Charge (Q) = 20000 c

Resistance (R) = 10 $\Ohm$

Time taken (t) = 10 seconds

$\rule{200}{1}$

★ To Find :

We have to find the work done

$\rule{200}{1}$

★ Solution :

We know that,

$\Large{\implies{\boxed{\boxed{\sf{I = \dfrac{Q}{t}}}}}}$

★ Putting Values ★

$\sf{\dashrightarrow I = \frac{2000 \cancel{0}}{\cancel{10}}} \\ \\ \sf{\dashrightarrow I = 2000}$

$\Large{\implies{\boxed{\boxed{\sf{I = 2000 \: A}}}}}$

$\rule{200}{2}$

Now,

$\Large{\implies{\boxed{\boxed{\sf{W = v \times I \times 2}}}}}$

$\sf{\dashrightarrow v = 10 \times 2000 \times 10} \\ \\ \sf{\dashrightarrow v = 200000}$

$\Large{\implies{\boxed{\boxed{\sf{W = 200000 \: J}}}}}$

Answer 2

Explanation:

$\huge\green{SoLu个i۝n:}$

$\red{V = 10v}$

$\blue{Q = 20,000 C}$

$\orange{R = 20 ohm}$

$\purple{t = 10sec}$

Equation to find out the work done can be derived from :

$v \: = \frac{w}{q}$

Here,

$\pink{V = volt}$

$\green{W =work done}$

$\blue{Q = charge}$

Hence,

$w \: = v \: \times q$

$10 \times 20000 \\ 200000$

Therefore,

$\blue{Q = 200000J}$

OR

$\red{Q = 2 メ 10^5J}$

HØPÊ IT HÈLPẞ YØÜ ☃️☄️