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Question
An alternating current I = 14 sin (100 πt) A passes through a series combination of a resistor of 30 Ω and an inductor of `(2/(5pi))` H. Taking `sqrt2` = 1.4 calculate the rms value of the voltage drops across the resistor and the inductor.
Solution
Given: I = 14 sin (100 πt) A, R = 30 Ω, L = `2/(5pi)`H
From here
ω = 100π
`X_L = omegaL` (inductive reactance)
= `100pi xx 2/(5pi) = 40Omega`
Impedance, Z = `sqrt(R^2 + X_L^2) = sqrt((30)^2 + (40)^2)`
Z = `sqrt(900 + 1600) = 50Omega`
`V_{rms} = I_{rms} xx Z`
We know `I_{rms} = I_0/sqrt2 = 14/1.4 = 10A`
So, `V_{rms} = 10 xx 50`
`V_{rms} = 10 xx 50`
`V_{rms}` = 500V
The voltage drop across the resistor,
`V_R = I_{rms} xx R`
= `10 xx 30`
= 300V
The voltage drop across the inductor,
`V_L = I_{rms} xx X_2`
= `10 xx 40`
= 400V
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