Answer to Question #315051 in Electric Circuits for Apc

Question #315051

In an R-L-C series circuit a maximum current of 0.5 A is obtained by varying the value of inductance L. The supply voltage is fixed at 230 V, 50 Hz. When maximum current flows through the circuit, the voltage measured across the capacitor is 350 V. What are the values of the cricuit parameters? 15

1
Expert's answer
2022-03-21T12:44:57-0400

We know that

Z=R2+(XLXC)2Z=\sqrt{R^2+(X_L-X_C)^2}

Maximum current

XL=XcX_L=X_c

Now impedance

Z=RZ=R

imax=VRi_{max}=\frac{V}{R}

R=2300.5=460ΩR=\frac{230}{0.5}=460\Omega

Now capacitor across Voltage

V=XLIV=X_L I

V=wLIV=w LI

L=V2πfIL=\frac{V}{2\pi f I}

L=2302×3.14×50×0.5=1.464HL=\frac{230}{2\times3.14\times50\times0.5}=1.464H

Critical condition

XL=XcwL=1wCX_L=X_c\\wL=\frac{1}{wC}

C=1w2LC=\frac{1}{w^2L}


C=1(2×3.14×50)2×1.464=6.92×106FC=\frac{1}{(2\times3.14\times50)^2\times 1.464}=6.92\times10^{-6}F

C=6.92μFC=6.92\mu F


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