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Question

Question: In the given ac circuit capacitance is adjustable, for what value of \( c \) , the current amplitude...

In the given ac circuit capacitance is adjustable, for what value of cc , the current amplitude is maximized?

(A) 4nF4nF
(B) 40nF40nF
(C) 4μF4\mu F
(D) 40μF40\mu F

Explanation

Solution

Hint : We can see that the given ac circuit is an RLC circuit. In an RLC circuit, the maximum amplitude of the current is during resonance. At resonance the inductive reactance and capacitive reactance are equal. Finding the values of both the reactance and equating them we can find the value of capacitance at which this is possible.

Formula used: Inductive reactance XL=ωL{X_L} = \omega L
Here, inductance is represented by LL , Angular frequency is represented by ω\omega
Capacitive reactance is represented by Xc=1ωc{X_c} = \dfrac{1}{{\omega c}}
Capacitance is represented by cc , Angular frequency is represented by ω\omega .

Complete step by step answer
We know that at resonance in an RLC circuit the impedance is minimum and the amplitude of the current is maximum. At this condition inductive reactance and capacitive reactance are equal.
From the question voltage is equal to v=100sin103tv = 100\sin {10^3}t
The general formula of voltage is v=vosinωtv = {v_o}\sin \omega t
Here ω\omega is the angular frequency 103{10^3}
We now know that angular frequency in the given ac circuit is 103{10^3}
Inductive reactance is equal to
XL=ωL\Rightarrow {X_L} = \omega L
XL=103×25×103=25\Rightarrow {X_L} = {10^3} \times 25 \times {10^{ - 3}} = 25
Capacitive reactance is equal to
Xc=1ωc\Rightarrow {X_c} = \dfrac{1}{{\omega c}}
Xc=1103×c\Rightarrow {X_c} = \dfrac{1}{{{{10}^3} \times c}}
At resonance inductive reactance and capacitive reactance are equal.
Xc=XL\Rightarrow {X_c} = {X_L}
1103×c=25\Rightarrow \dfrac{1}{{{{10}^3} \times c}} = 25
Solving for capacitance we get
1c=25×103\Rightarrow\dfrac{1}{c} = 25 \times {10^3}
c=125×103=40×106=40μF\Rightarrow c = \dfrac{1}{{25 \times {{10}^3}}} = 40 \times {10^{ - 6}} = 40\mu F
Hence the value of capacitance is 40μF40\mu F
Option (D) 40μF40\mu F is the correct answer .

Note
We can say that the amplitude of the current is maximized during resonance because at resonance the impedance of the circuit is minimum which means the current is maximum. Because the current is alternating current we can say that at maximum current the amplitude will be maximum. The units of capacitance are Farad. The frequency of current and voltage at which resonance takes place is called resonant frequency.