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A resistor, an inductor and a capacitor ...

A resistor, an inductor and a capacitor are connected in series with a 120 V, 100 Hz ae source. Voltage leads the current by `35^(@)` in the circuit. If the resistance of the resistor is `10Omega` and the sum of inductive and capacitive reactances is `17Omega`, calculate the self-inductance of the inductor.

Text Solution

Verified by Experts

Given `R=10Omega`
`X_(L)+X_(C)=17Omega`
given `phi=35^(@)`
L = ?
We know that `tanphi=(X_(L)-X_(C))/R`
Hence `X_(L)-X_(C)=Rtanphi`
= `10xxtan35^(@)`
`10xx0.7002`
= 7.002
i.e., `X_(L)-X_(C)~~7Omega`
Since `X_(L)+X_(C)=17`
and `X_(L)-X_(C)=7`
Solving we get `2X_(L)=24`
`thereforeX_(L)=12Omega`
But `X_(L)=2pi"fL=12"`
`thereforeL=12/(2xx3.142xx100)=0.0190H`
Self inductance of the coil = 0.019H
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