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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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Answer» 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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