
A capacitor, an inductor and an electric bulb are connected in series to an AC supply of variable frequency. As the frequency of the supply is increased gradually, then the electric bulb is found to:
(A) Increase in brightness
(B) Decrease in brightness
(C) Increase, reach a maximum and then decrease in brightness
(D) Show no change in brightness
Answer
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Hint: The supply connected is AC. We know in AC it changes its path after every half cycle and for a complete cycle its displacement is zero. For an inductor when AC flows through it, an induced emf is induced and it gets build up. The inductive reactance is the opposition to the flow of AC through it and is directly proportional to the frequency of the AC. Also, when AC passes through the capacitor it starts charging up and when it is fully charged no AC can pass through it.
Complete step by step answer:
The capacitor, the inductor and the electric bulb which can be considered as a resistor are all connected in series combination. The impedance is the net reactance offered to the flow of current and is given by,
$Z=\sqrt{{{R}^{2}}+{{(2\pi vL-\dfrac{1}{2\pi vC})}^{2}}}$
where R is the resistance of the electric bulb, L is the impedance of the inductor, C is the capacitance of the capacitor and $v$ is the frequency of the AC.
So, current can be given as:
$I=\dfrac{V}{Z=\sqrt{{{R}^{2}}+{{(2\pi vL-\dfrac{1}{2\pi vC})}^{2}}}}$
It is clear from the expression that as frequency increases the term in the bracket in the denominator decreases first then it starts increasing. So, the brightness increases first and then decreases.
So, the correct option is C.
Note: When we have a series LCR circuit, then the total opposition to the flow of the current is calculated by adding the resistance offered by the capacitor, the resistance and the inductor and that is known as impedance of the circuit. Quality factor related the energy stored in the circuit to the energy dissipated in each cycle of oscillation. Also, we can always find out the value of resonant frequency at which the current is maximum in the circuit.
Complete step by step answer:
The capacitor, the inductor and the electric bulb which can be considered as a resistor are all connected in series combination. The impedance is the net reactance offered to the flow of current and is given by,
$Z=\sqrt{{{R}^{2}}+{{(2\pi vL-\dfrac{1}{2\pi vC})}^{2}}}$
where R is the resistance of the electric bulb, L is the impedance of the inductor, C is the capacitance of the capacitor and $v$ is the frequency of the AC.
So, current can be given as:
$I=\dfrac{V}{Z=\sqrt{{{R}^{2}}+{{(2\pi vL-\dfrac{1}{2\pi vC})}^{2}}}}$
It is clear from the expression that as frequency increases the term in the bracket in the denominator decreases first then it starts increasing. So, the brightness increases first and then decreases.
So, the correct option is C.
Note: When we have a series LCR circuit, then the total opposition to the flow of the current is calculated by adding the resistance offered by the capacitor, the resistance and the inductor and that is known as impedance of the circuit. Quality factor related the energy stored in the circuit to the energy dissipated in each cycle of oscillation. Also, we can always find out the value of resonant frequency at which the current is maximum in the circuit.
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