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Question: A RC series circuit of \( R = 15\Omega \) and \( c = 10\mu F \) is connected to \( 20volt \) DC supp...

A RC series circuit of R=15ΩR = 15\Omega and c=10μFc = 10\mu F is connected to 20volt20volt DC supply for a very long time. Then the capacitor is disconnected from the circuit and connected to the inductor 10mH10mH . Find the amplitude of the current.
(A) 0.210A\left( A \right){\text{ 0}}{\text{.2}}\sqrt {10} A
(B) 210A\left( B \right){\text{ 2}}\sqrt {10} A
(C) 0.2A\left( C \right){\text{ 0}}{\text{.2A}}
(D) 10A\left( D \right){\text{ }}\sqrt {10} A

Explanation

Solution

To solve this we need the formula of the amplitude of the current and is given by I∘=Q∘w{I_ \circ } = {Q_ \circ }w and we also know that the charge is given by, Q∘=CV∘{Q_ \circ } = C{V_ \circ } , so by using all these formulae we will be able to solve this question.

Formula used:
The amplitude of the current is given by,
I∘=Q∘w{I_ \circ } = {Q_ \circ }w
Here, I∘{I_ \circ } , will be the amplitude current
Q∘{Q_ \circ } , will be the charge
ww , will be the angular frequency
The charge is given by
Q∘=CV∘{Q_ \circ } = C{V_ \circ }
Here, CC will be the capacitor
Angular frequency is given by,
w=1LCw = \dfrac{1}{{\sqrt {LC} }} .

Complete step by step solution:
Here in this question we have to find the amplitude of the current. For this we have the values as a capacitor which is given as c=10μFc = 10\mu F and similarly the resistor is given by R=15ΩR = 15\Omega .
So for solving, as we know that Impedance of current will be given by
⇒I∘=Q∘w\Rightarrow {I_ \circ } = {Q_ \circ }w
Now on substituting the value of charge which is given in the formula, we will get
⇒I∘=CV∘w\Rightarrow {I_ \circ } = C{V_ \circ }w
And from this angular frequency can also be written as
⇒I∘=CV∘×1LC\Rightarrow {I_ \circ } = C{V_ \circ } \times \dfrac{1}{{\sqrt {LC} }}
On solving furthermore, we will get the equation as
⇒I∘=CLV∘\Rightarrow {I_ \circ } = \sqrt {\dfrac{C}{L}} {V_ \circ }
So now on substituting the values, we will get the equation as
⇒I∘=10×10−6010×10−3×20\Rightarrow {I_ \circ } = \sqrt {\dfrac{{10 \times {{10}^{ - 60}}}}{{10 \times {{10}^{ - 3}}}}} \times 20
Since the liker term will cancel each other so on solving it we will get the equation as
⇒I∘=0.210A\Rightarrow {I_ \circ } = 0.2\sqrt {10} A
Therefore, the amplitude of the current will be 0.210A0.2\sqrt {10} A .
Hence, the option (A)\left( A \right) is correct.

Note:
So an RC series circuit having both the capacitor and the resistor. And a capacitor can accumulate the energy and a resistor positioned in the series will switch the rate at which it charges or discharges. And the characteristic time dependence will be exponential.