Question

11. A capacitor of capacitance 10^-4/pi F, an inductor of inductance 2/ pi H and a resistor of resistance 100 ohm are connected to form a series RLC circuit. When an AC supply of 220V, 50 Hz is applied to the circuit, determine the impedance of the circuit.​

Answer 1

Answer:

141.4 ohms

Explanation:

Attached above

Answer 2

Given :

Capacitance (C) = $\frac{10^{-4} }{\pi }$ F

Inductance (L) = $\frac{2}{\pi }$ H

Resistance (R) = 100 ohm

Frequency (n) = 50 Hz

To Find :

Impedance of the circuit

Solution :

Angular frequency (ω) = 2$\pi$n

                                     = 2$\pi$50 = 100

Inductive reactance ($X_{L}$) = Lω = $\frac{2}{\pi }$ × 100$\pi$

                                         = 200 ohm

Capacitive reactance ($X_{c}$) = $\frac{1}{Cw}$ = $\frac{1}{\frac{10^{-4} }{\pi } * 100\pi }$

                                           = $10^{2}$ = 100 ohm

Impedance (z) = $\sqrt{(X_{L} - X_{C})^{2} + R^{2}$

                    z  =  $\sqrt{(200-100 )^2 + 100^{2}$

                    z  =  $\sqrt{100^{2} + 100^{2} }$

                    z  = $\sqrt{100^{2}(1+1) }$

                    z  = 100$\sqrt{2}$

                    z  = 141.4 ohm

∴ The impedance of the given LCR circuit is 141.4 ohm.