Answer to Question #86731 in Electric Circuits for Adam

Question #86731
A resistance R and inductance L=0.01 H and a capacitance are connected in series when a voltage V=400Cos (300t-10) v is applied to the series combination, the current flowing I=10 √2cos (300t-55°)A. Find P and C.
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Expert's answer
2019-03-21T16:16:52-0400

The active power:


P=12VIcos(ϕVϕI)=P=\frac{1}{2}VI\text{cos}(\phi_V-\phi_I)==12400102cos(10(55))=2000 W.=\frac{1}{2}\cdot400\cdot10\sqrt{2}\text{cos}(-10^\circ-(-55^\circ))=2000\text{ W}.

Find the resistance:


R=PI2=2000102=20 Ω.R=\frac{P}{I^2}=\frac{2000}{10^2}=20\space\Omega.

Write Ohm's law for the circuit:


Z=R2+(ωL1/ωC)2=VI,Z=\sqrt{R^2+(\omega L-1/\omega C)^2}=\frac{V}{I},

202+(3000.011300C)2=(400102)2,20^2+(300\cdot0.01-\frac{1}{300 C})^2=(\frac{400}{10\sqrt{2}})^2,

C=144.93 μFC=144.93\space\mu\text{F}


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