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Electrostatic Potential And Capacitance

Question
CBSEENPH12039372

(i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase: current or voltage?

(ii) Without making any other change, find the value of the additional capacitor C, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity.

Solution

Given V = Vosin (1000 +Φ)
ω = 1000s-1
L = 100 mH
C =2μF
R = 400Ω
Phase difference  straight capital phi space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator straight X subscript straight L minus straight X subscript straight C over denominator straight R end fraction close parentheses
XL = ωL = 1000 x100 x10-3 = 100Ω
 =straight X subscript straight C space equals space 1 over ωC space equals space fraction numerator 1 over denominator 1000 space straight x 2 space straight x 10 to the power of negative 6 end exponent end fraction space equals space 500 space straight capital omega
straight capital phi space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 100 minus 500 over denominator 400 end fraction close parentheses space equals space tan to the power of negative 1 end exponent left parenthesis negative 1 right parenthesis
straight capital phi space equals space minus space 45 to the power of straight o space and space the space current space is space leading space the space voltage.

(ii) For power factor to be unity, R = Z
or XL = XC
straight omega squared space equals space 1 over LC space left parenthesis straight C space equals space resultant space capacitance right parenthesis
10 to the power of 6 space equals space fraction numerator 1 over denominator 100 space straight x 10 to the power of negative 3 end exponent space xC end fraction
rightwards double arrow space straight C to the power of apostrophe space equals space 10 to the power of negative 5 end exponent space straight F

For two capacitance in parallel, resultant capacitance C'=C + C1
10-5 = 0.2 x 10-5 + C1
⇒ C1 = 8μF

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