Question

At a moment (t=0) when charge on capacitor c1 is zero, the switch is closed. if i0 be the current through inductor at that instant, for t 0, find the maximum charge on c1

Answer 1

Explanation:

since at t = 0, charge is maximum (=q0)

Therefore current will be zero.

21Li2=31 (21cq2)

or i=31cq=3qw

From the expression i=wq02-q2

we have, 3qw=wq02-q2

or q=23q0

Since at t=0, charge in maximum or q0 so we can write

q =

23

24

2689

hope it's helps u...

Answer 2

Given,

t = 0 when the charge on C₁ = 0

I₀ = current through the indicator.

To find,

Maximum charge on C₁.

Solution,

The situation is given as C₁ and C₂ are capacitors and L is the inductor.

Using the formula of maximum energy stored in the inductor and capacitor,

U₁ = $\frac{1}{2}$ LI₀² ....(i)

Here I₀ is the maximum current.

The energy stored in the capacitor for the maximum potential across the capacitor is

U₂ = $\frac{1}{2}$ CV² ....(ii)

Here C is the equivalent capacitance and V is the potential across the capacitor.

The equivalent capacitance of the parallel capacitors C₁ and C₂ is

C =  C₁ + C₂ ....(iii)

The charge stored in the capacitor is defined as

Q=CV ....(iv)

According to the conservation of energy,\

      U₁ = U₂

=> $\frac{1}{2}$ LI₀² =  $\frac{1}{2}$ CV²

Substituting C from equation (iii)

$\frac{1}{2}$ LI₀² =  $\frac{1}{2}$ (C₁ + C₂)V²

Using equation (iv)

$Q=\sqrt[C^1]{\frac{ LI_0^2}{C_1+C_2} }$

$Q=\sqrt[I_0C^1]{\frac{ L}{C_1+C_2} }$

Hence, the maximum charge on C₁= $\sqrt[I_0C^1]{\frac{ L}{C_1+C_2} }$

#SPJ3