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.

Answer 1

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