An AC source of frequency 50 Hz is connected to 50 mH inductor and a bulb. To get the maximum brightness from the bulb, what should be the capacitance of a capacitor connected in series with the circuit?

Question

An AC source of frequency 50 Hz is connected to 50 mH inductor and a bulb. To get the maximum brightness from the bulb, what should be the capacitance of a capacitor connected in series with the circuit? (A)  50× 10-4F  (B)  1× 10-4F  (C)  2.03× 10-4F  (D)  3× 10-4F

Tardigrade Admin · Accepted Answer

For maximum brightness maximum current is required in the circuit. To get maximum current, XL=XC (at resonance) ⇒ 2π vL=(1/2π vC) ⇒ C=(1/4π 2v2L) =(1/4(3.14)2(50)2(50× 10-3)) =2.03× 10-4F

Q. An AC source of frequency 50 Hz is connected to 50 mH inductor and a bulb. To get the maximum brightness from the bulb, what should be the capacitance of a capacitor connected in series with the circuit?

$50 \times 10^{- 4} F$

$1 \times 10^{- 4} F$

$2.03 \times 10^{- 4} F$

$3 \times 10^{- 4} F$

Q. An AC source of frequency 50 Hz is connected to 50 mH inductor and a bulb. To get the maximum brightness from the bulb, what should be the capacitance of a capacitor connected in series with the circuit?

50×10−4F

1×10−4F

2.03×10−4F

3×10−4F

For maximum brightness maximum current is required in the circuit. To get maximum current, XL​=XC​ (at resonance) ⇒ 2πvL=2πvC1​ ⇒ C=4π2v2L1​ =4(3.14)2(50)2(50×10−3)1​ =2.03×10−4F

$50 \times 10^{- 4} F$

$1 \times 10^{- 4} F$

$2.03 \times 10^{- 4} F$

$3 \times 10^{- 4} F$