Question

A 120 V, 20 Hz source is connected to a series circuit consisting of 5 ohm capacitive reactance, a 1.6 ohm resistor
and a coil with resistance and inductive reactance of 3-ohm and 1.2-ohm respectively. Calculate (a) voltage
across the coil (b) the resonant frequency.

Answer 1

Answer:

Explanation:

a) inducive reactance ,XL =1.2 ohm

       impedance z =$\sqrt{R^{2} + XL^{2} }$.

      ie, z= $\sqrt{3^{2} +1.2^{2} }$ = 3.2 ohm.

      current i =total voltage/ z.

=>  I = 120/3.2 =37.5 ohm.

so voltage across the coil = R x I = 3 x 37.5 = 112.5 V.

b)   XL = 1 / 2*pi*F*c

            therefore, C = 1 / 2* 3.14* 20* 1.2 =>6.6  F

 L = XL /2 * pi * F  =  9.5 ( at resonance point XL =XC )

therefore resonant frequency = 1 / 2* pi* sq (L * C) .

              =>1 / 2 * 3.14* $\sqrt{9.5*6.6}$  = 0.020 HZ.