A 400 Ω resistor, a 3 H inductor and a 5 μF capacitor are connected in series to a 220 V, 50 Hz ac source. Calculate the: (1) Impedance of the circuit…

Given: R=400 Ω, L=3 H, C=5 μF, V=220 V, f=50 Hz. (1) X_L=2π fL=2π×50×3=942.5 Ω; X_C=1/(2π fC)=1/(2π×50×5×10⁻⁶)=636.6 Ω. Z=√(R²+(X_L-X_C)²)=√(400²+305.9²)=503.5 Ω. (2) I=V/Z=(220)/(503.5)=0.437 A. Z = 503.5 Ω, I = 0.437 A

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Solve the following.

A 400 Ω resistor, a 3 H inductor and a 5 μF capacitor are connected in series to a 220 V, 50 Hz ac…

(i)[3]

A $400\ \Omega$ resistor, a $3\text{ H}$ inductor and a $5\ \mu\text{F}$ capacitor are connected in series to a $220\text{ V}, 50\text{ Hz}$ ac source. Calculate the: (1) Impedance of the circuit. (2) Current flowing through the circuit.

(ii)[2]

Draw a labelled graph showing the variation of impedance ($Z$) of a series LCR circuit versus frequency ($f$) of the ac supply.

Draw: Graph of impedance (Z) versus frequency (f) for a series LCR circuit

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Answer

Answer (i)

Formulas used: Series LCR circuit from L, C and f: Z = sqrt(R**2 + (XL - XC)**2) with C = 5e-6 F, L = 3 H, R = 400 ohm, f = 50 Hz, so Z = 503.5 ohm Ohm's law: V = I*R with R = 503.5 ohm, V = 220 V, so I = 0.437 A

Given: $R=400\ \Omega$, $L=3\ \text{H}$, $C=5\ \mu\text{F}$, $V=220\ \text{V}$, $f=50\ \text{Hz}$. (1) $X_L=2\pi fL=2\pi\times50\times3=942.5\ \Omega$; $X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi\times50\times5\times10^{-6}}=636.6\ \Omega$. $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{400^2+305.9^2}=503.5\ \Omega$. (2) $I=\frac{V}{Z}=\frac{220}{503.5}=0.437\ \text{A}$.

Final answer: $Z = 503.5\ \Omega,\ I = 0.437\ \text{A}$

Answer (ii)

$Z$ falls as $f$ increases, reaches its minimum value $Z=R$ at the resonant frequency $f_0=\frac{1}{2\pi\sqrt{LC}}$ (where $X_L=X_C$), and then rises again.