Question

: A Capacitor of Capacitance 8micro farad is charged upto 1000v. then find its stored energy?


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Answer 1

Answer:

Appropriate Question :-

• A capacitor of capacitance 8μF is charged to a potential difference of 1000 V. What is the storage energy.

Given :-

• A capacitor of capacitance 8μF is charged to a potential difference of 1000 V.

To Find :-

• What is the storage energy.

Formula Used :

$\mapsto \sf\boxed{\bold{\pink{E =\: \dfrac{1}{2}CV^2}}}$

where,

• E = Energy
• C = Capacitance
• V = Potential Difference

Solution :-

First, we have to convert the capacitance μF into F :

$\implies \sf Capacitance =\: 8{\mu}F$

$\implies \sf Capacitance =\: 8 \times 10^{- 6}\: F\: \: \bigg\lgroup \sf\bold{\pink{1{\mu}F =\: 10^{- 6}\: F}}\bigg\rgroup$

$\implies \sf\bold{\purple{Capacitance =\: 8 \times 10^{- 6}\: F}}$

Now, we have to find the stored energy :

Given :

• Capacitance (C) = 8 × 10-⁶ F
• Potential Difference (V) = 1000 V

According to the question by using the formula we get,

$\longrightarrow \sf E =\: \dfrac{1}{2} \times 8 \times 10^{- 6} \times (1000)^2$

$\longrightarrow \sf E =\: \dfrac{1}{2} \times 8 \times 10^{- 6} \times 1000 \times 1000$

$\longrightarrow \sf E =\: \dfrac{1}{2} \times 8 \times 10^{- 6} \times 1000000$

$\longrightarrow \sf E =\: \dfrac{1}{2} \times 8 \times 10^{- 6} \times 10^6$

$\longrightarrow \sf E =\: \dfrac{1}{2} \times 8 \times 10^{(- 6 + 6)}$

$\longrightarrow \sf E =\: \dfrac{1}{2} \times 8 \times 10^0$

$\longrightarrow \sf E =\: \dfrac{1}{2} \times 8 \times 1$

$\longrightarrow \sf E =\: \dfrac{1}{\cancel{2}} \times {\cancel{8}}$

$\longrightarrow \sf\bold{\red{E =\: 4\: J}}$

${\small{\bold{\underline{\therefore\: The\: stored\: energy\: is\: 4\: J\: .}}}}$