Find the voltage across the various elements, i.e., resistance, capacitance and inductance which are in series and having values `1000 Omega, 1muF` an

Monjur Alahi

Find the voltage across the various elements, i.e., resistance, capacitance and inductance which are in series and having values `1000 Omega, 1muF` and 2.0 H respectively . Given emf as,
V = 100sqrt2 sin 1000 t V`


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Salim Ahmed Kawsar

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The rms value of voltage across the source, `V_(rms) = V_(0)/sqrt(2)`
comparing the given equation with general equation,
`V_(0) = 100sqrt(2),omega=1000rads^(-1)`
Also, R=`1000Omega`, C=`1muF= 1xx10^(-6)`F
L=2H
`V_(rms)=(100sqrt(2))/(sqrt(2)) = 100V`
`therefore I_(rms) = V_(rms)/|Z| = (V_(rms))/(sqrt(R^(2) + (X_(L)-X_(C))^(2)))` = `V_(rms)/(sqrt(R^(2)+(omegaL-1/(omegaC))^(2)))`
`=100/sqrt((1000)^(2)+(1000xx2-1/(1000 xx 1xx 10^(-6))))`
The current will be same everywhere in the circuit, therefore, PD across resistor,
`V_(R) = I_(rms)R=0.0707xx1000 = 70.7V`
PD across inductor
`V_(L) = I_(rms)X_(L) = 0.0707 xx 1/(1xx1000xx10^(-6)) = 70.7` V
Note:" The rms voltages do not add directly as, `V_(R)+V_(L)+V_(C)=282.8`V
Which is not the source voltage 100V. The reason is that these voltages are not in phase and cna be added by vector or by phasor algebra.

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