Question

in lc oscillation maximum charge on capacitor is q the current in the circuit when 50 percent energy is electrical and 50 percent magnetic is​

Answer 1

Answer:

current in the circuit when 50% energy is stored in the form of electrical energy is given as

$I = \sqrt{\frac{q^2}{2LC}}$

Explanation:

As we know that maximum charge on the capacitor is q

so maximum total energy stored in the system is given as

$U = \frac{1}{2}CV^2$

$U = \frac{q^2}{2C}$

now we know that total energy is always conserved in LC oscillation

so we have

$U = \frac{1}{2}LI^2 + \frac{Q^2}{2C}$

now it is given that 50% energy is stored in form of electrical energy and other 50% is in form of magnetic energy

so we have

$\frac{Q^2}{2C} = \frac{1}{2}LI^2$

so we will have

$\frac{q^2}{2C} = 2(\frac{1]{2}LI^2)$

$\frac{q^2}{2C} = LI^2$

$I = \sqrt{\frac{q^2}{2LC}}$

#Learn

Topic : LC oscillation

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