Question

11: The voltage of a capacitor 12F with a rating of 2J
energy is
2499742659
0.57
5.7
57
570​

Answer 1

It has given that, a capacitor of capacitance 12 F which is connected in a circuit of voltage v, energy conserved in the capacitor is 2 J.

We have to find the voltage v.

Solution : we know, Energy conserved in a capacitor is given by, U = 1/2 Cv²

Here C = capacitance of capacitor = 12 F

Energy stored, U = 2 J

So, 2 J = 1/2 × 12 × V

⇒(1/3) = V²

⇒V = 0.57 volt

Therefore voltage is 0.57 volt.

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Answer 2

GiveN :-

• A capacitor of capacitence = 12F Which is connected in a circuit of Voltage 'V'

• Energy comserved in the capacitor = 2J

To FinD :-

• Enrgy ( V )

CalculaTioN :-

Energy Conserved in a Capacitor

$\bold{u \: = \frac{1}{2} \times C {v}^{2} }{}$

Where,

'U' = Capacitence of the Capasitor

'V' = Energy Stored = 2J

So,

$2j = \frac{1}{2} \times 12 \times v \\ \\ \bold{by \: simplification}{} \\ \\ \frac{1}{3} = {v}^{2} \\ \\ \bold{ \red{★Hence \: v = 0.57v}{} }{}$

Hence Voltage = 0.57 Volt

$\boxed{ \bold{ \blue{★Option \: A}{} }{} }{}$