Question

In series RL circuit, R = 5 Ω, XL = 10 Ω and Xc = 15 Ω If this circuit is connected to a voltage source v = 100 sin (314t + 30) V, the rms value of the current will be...​

Answer 1

Given:

A series RLC circuit with

$R=5$Ω

$X_L=10$Ω

$X_C=15$Ω

is connected across a voltage source, $V=100sin(314t+30) V$

To Find:

RMS value of the current.

Solution:

It is given that in the RLC series circuit, $R=5$Ω,  $X_L=10$Ω, and $X_C=15$Ω.

Impedance in an RLC series circuit, $Z=\sqrt{R^2+(X_C-X_L)^2}$

Using the formula, we obtain $Z=\sqrt{(5)^2+(-5)^2}$

                                                    $=5\sqrt{2}$Ω.

Expression of voltage given, $V=100sin(314t+30) V$

⇒ $V_o=100V$ where $V_o$ is the maximum voltage.

By definition, $V_{rms}=\frac{V_o}{\sqrt{2} }$

                  ⇒ $V_{rms}=50\sqrt{2}$

Using Ohm's law, $I_{rms}=\frac{V_{rms}}{Z}$

                                     $=\frac{50\sqrt{2} }{5\sqrt{2} }$

                                     $=10A$.

Hence, the RMS value of the current in this circuit is equal to 10A.