
In an LC circuit the capacitor has maximum charge ${q_0}$. The value of ${\left( {\dfrac{{dI}}{{dt}}} \right)_{\max }}$, where I is the current in the circuit and $t$ is time, is
$\left( A \right)\dfrac{{{q_o}}}{{LC}}$
$\left( B \right)\dfrac{{{q_o}}}{{\sqrt {LC} }}$
$\left( C \right)\dfrac{{{q_o}}}{{LC}} - 1$
$\left( D \right)\dfrac{{{q_o}}}{{LC}} + 1$
Answer
594.6k+ views
Hint: In this question the connection is series and in series connection current always remains the same, so according to the maximum power transfer theorem the power on both the components should be equal therefore voltage of both the components should be equal so use these concepts to reach the solution of the question.
Formula used – $\left( {{V_L}} \right) = L\left( {\dfrac{{dI}}{{dt}}} \right)$, $\left( {{V_C}} \right) = \dfrac{Q}{C}$
Complete Step-by-Step solution:
As we see that L and C are connected in series so for maximum case the voltage on both should be equal.
Let the voltage on inductor (L) = ${V_L}$ and voltage on capacitor (C) = ${V_C}$
$ \Rightarrow {V_L} = {V_C}$
$ \Rightarrow {\left( {{V_L}} \right)_{\max }} = {\left( {{V_C}} \right)_{\max }}$................ (1)
Now as we know that in series connection current remains the same.
So the voltage on inductor is
$ \Rightarrow {\left( {{V_L}} \right)_{\max }} = L{\left( {\dfrac{{dI}}{{dt}}} \right)_{\max }}$
And the voltage on capacitor is
$ \Rightarrow {\left( {{V_C}} \right)_{\max }} = \dfrac{{{Q_{\max }}}}{C}$
Now it is given that ${Q_{\max }} = {q_o}$
$ \Rightarrow {\left( {{V_C}} \right)_{\max }} = \dfrac{{{q_o}}}{C}$
Now according to equation (1) equate these two equations we have,
$ \Rightarrow L{\left( {\dfrac{{dI}}{{dt}}} \right)_{\max }} = \dfrac{{{q_o}}}{C}$
$ \Rightarrow {\left( {\dfrac{{dI}}{{dt}}} \right)_{\max }} = \dfrac{{{q_o}}}{{LC}}$
So this is the required answer.
Hence option (A) is the correct answer.
Note – Whenever we face such types of question the key concept is series connection as in series connection current remains same and we have to developed the condition for maximum phenomenon so that voltage on both the components should be equal so simplify equate them and write component general voltage formula as above and simplify we will get the required answer.
If the connection is in parallel, voltage remains the same but passing current from each component is different.
Formula used – $\left( {{V_L}} \right) = L\left( {\dfrac{{dI}}{{dt}}} \right)$, $\left( {{V_C}} \right) = \dfrac{Q}{C}$
Complete Step-by-Step solution:
As we see that L and C are connected in series so for maximum case the voltage on both should be equal.
Let the voltage on inductor (L) = ${V_L}$ and voltage on capacitor (C) = ${V_C}$
$ \Rightarrow {V_L} = {V_C}$
$ \Rightarrow {\left( {{V_L}} \right)_{\max }} = {\left( {{V_C}} \right)_{\max }}$................ (1)
Now as we know that in series connection current remains the same.
So the voltage on inductor is
$ \Rightarrow {\left( {{V_L}} \right)_{\max }} = L{\left( {\dfrac{{dI}}{{dt}}} \right)_{\max }}$
And the voltage on capacitor is
$ \Rightarrow {\left( {{V_C}} \right)_{\max }} = \dfrac{{{Q_{\max }}}}{C}$
Now it is given that ${Q_{\max }} = {q_o}$
$ \Rightarrow {\left( {{V_C}} \right)_{\max }} = \dfrac{{{q_o}}}{C}$
Now according to equation (1) equate these two equations we have,
$ \Rightarrow L{\left( {\dfrac{{dI}}{{dt}}} \right)_{\max }} = \dfrac{{{q_o}}}{C}$
$ \Rightarrow {\left( {\dfrac{{dI}}{{dt}}} \right)_{\max }} = \dfrac{{{q_o}}}{{LC}}$
So this is the required answer.
Hence option (A) is the correct answer.
Note – Whenever we face such types of question the key concept is series connection as in series connection current remains same and we have to developed the condition for maximum phenomenon so that voltage on both the components should be equal so simplify equate them and write component general voltage formula as above and simplify we will get the required answer.
If the connection is in parallel, voltage remains the same but passing current from each component is different.
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