Circuit question of inductor and resistor?

Question

A 1.15 kilo Ohms resistor and a 590 mH inductor are connected in series to a 1350 Hz generator with an rms voltage of 12.1 V.

What is the rms current in the circuit in mA?

What capacitance must be inserted in series with the resistor and inductor to reduce the rms current to half the value found in part A in nF?

DH · Accepted Answer

A) the reactance of the inductor is XL = 2π*f*L = 2π*1350*0.590 =5005Ω

So the impedance Z = sqrt(1150^2 + 5005^2) = 5135Ω

Therefore the rms current = V/Z = 12.1V/5135Ω = 2.36x10^-3A = 2.36 mA

B) If the current is reduced to 1/2 then then impedance must double

So Z = sqrt(R^2 + (XlL- XC)^2) = 10270

So 1150^2 + (5005 - XC)^2 = 10270^2

or 5005 - XC = +-sqrt(10270^2 - 1150^2) = +-10205

or XC = 5005 + 10205 = 15210

Now XC = 1/2πf*C

or C = 1/(2π*1350*15210) = 7.75x10^-9F = 7.75nF

Trending News

Circuit question of inductor and resistor?

1 Answer