Question

A 55.4 ? resistor is connected in series with a 16.2 microF capacitor and a 60 Hz, 140 V source.

a) Find the current in the circuit. Answer in units of mA.

b) What is value of the inductor that must be inserted in the circuit to reduce the current to one half the previous current?

Answer in units of H.

Answer 1

The capacitive reactance is,

           XC = 1/C = 1/2fC = 1/2(60)(16.2x10^-6) = 165.786

The impidence is,

         Z = [X_C² + R²]^1/2 = [(165.786)² + (55.4)²]^1/2 = 174.7978

The current in the circuit is,

         I = V/Z = 140 / 174.7978 = 0.8 A = 800 mA

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The new current is, I'= I /2 =0.8 /2 = 0.4 A

So, the current is,

          I' = V / [R² + [X_L - X_C]²]^1/2

         0.4 = 140 / [ [ (55.4)²+[XL -165.786]²]^1/2

Solve for X_L, and we get X_L = 511.373

Hence, the inductance of the inductor is,

        X_L = 2fL

        L = X_L/2f = 511.373 / 2()(60) = 1.356 H