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enter image description here This example is given in my book as I know voltage of the source in an RLC circuit can be obtained using following formula $$u=U_m*sin (wt+Φ)$$ But in this example the voltage of source is given in the form of $$u=U_m*sinwt$$ Why is this so?

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It is often the case that in such a series circuit the phase of the current is taken to be zero, $\Phi = 0$, because the current is the same in each part of the circuit.
This means that all voltages have a phase which is referenced to the phase of the current.

However in your example the supply voltage is chosen as having a phase angle of zero and so the phase of all other voltages and the current in the circuit are referenced to the supply voltage.

Using the current as the reference if you found the supply voltage to have a phase of $+30^\circ$ which means that the supply voltage leds the current by $30^\circ$.
On the other hand using the supply voltage as the reference in the same circuit you would find that the phase of the current was $-30^\circ$ meaning that the current lagged the supply voltage by $30^\circ$.


Update as a result of a comment from @Clair

The phasor diagrams for this circuit are shown below.

enter image description here

Since the reactance of the inductor $X_{\rm L} = 31.4 \, \Omega $ is larger than that of the capacitor $X_{\rm C} = 21.2 \, \Omega $ the supply voltage $V_{\rm supply}$ leads the current $I$ by $26.8^\circ$ or you could say that the current lags the supply voltage by $26.8^\circ$.

In terms of equations with $V_{\rm supply} = V_{\rm m} \sin (\omega t)$, ie the phase is referenced to the supply voltage, you would write that the current $I = I_{\rm m} \sin (\omega t - 26.8^\circ$) whereas if you took the current to be the reference as $I = I_{\rm m} \sin( \omega t)$ then the supply voltage is $V_{\rm supply} = V_{\rm m} \sin (\omega t + 26.8^\circ)$.

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  • $\begingroup$ But then the phase angle is found to be $26.6°$ not zero $\endgroup$ – Clair Feb 28 '18 at 6:11
  • $\begingroup$ @Clair That is the phase angle between the supply voltage and the current in the circuit (and also the voltage across the resistor). $\endgroup$ – Farcher Feb 28 '18 at 8:10
  • $\begingroup$ So why it is not given in the equation but it is given to be zero? $\endgroup$ – Clair Feb 28 '18 at 19:20
  • $\begingroup$ @Clair I do not understand your last comment. $\endgroup$ – Farcher Mar 1 '18 at 12:38
  • $\begingroup$ $U=200sin1000*pi*t$ in this equation phi is given to be zero but after calculation it is find to be $26.6°$ $\endgroup$ – Clair Mar 1 '18 at 14:27

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