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What is the value of inductance \[L\] for which the current is a maximum in a series LCR circuit with \[C = 10\mu F\] and \[\omega = 1000rad{\sec ^{ - 1}}\]?
A) \[10mH\]
B) \[100mH\]
C) \[1mH\]
D) Cannot be calculated unless R is known.

Last updated date: 15th Jun 2024
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Hint: In the question it is written that the current is maximum in the given series LCR circuit. This condition is true, at resonance. Series LCR circuit is also known as a resonant circuit or RLC circuit. Thus, we need to consider the relation between \[\omega \] \[L\] and\[C\], in order to get the value of \[L\].

Complete step by step solution:
We know that this circuit consists of inductance, resistance and capacitance in series or parallel.
We are aware that an electrical circuit is in resonance, when maximum current flows through the circuit for a particular source of alternating supply having a constant frequency.

At resonance, the capacitive reactance always equals to inductive reaction.
We know,
Capacitive reactance is formulated as:
\[{X_C} = \dfrac{1}{{\omega C}}\]
\[C\]is the capacitance and
\[\omega \] is the Resonant frequency
Inductive reactance is formulated as:
\[{X_L} = \omega L\]
\[L\]is the inductance.
\[\omega \] is the Resonant frequency
Thus, equating the above equations, we obtain:
\[\omega L = \dfrac{1}{{\omega C}}\]
Rearranging the equations, we get:
\[L = \dfrac{1}{{{\omega ^2}C}}\]
Now, putting the values, we obtain:
\[L = \dfrac{1}{{{{(1000)}^2}(10 \times {{10}^{ - 6}})}}\]

We multiplied the capacitance value with \[{10^{ - 6}}\] to get the value in Farad, which is the SI unit of capacitance.
\[L = 0.1H = 100mH\]
This is the required answer.

Therefore, option (B) is correct.

Note: An important point to be noted is that, here, in case of inductor, voltage leads current by an angle of \[{90^o}\], in case of capacitance, current leads voltage by an angle of \[{90^o}\] but in case of resistance, both current and voltage are in same phase. The units of all parameters must be considered in their respective SI units, otherwise it will give erroneous results. RLC circuits are used in tuning applications.