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MODEL PAPER 10 — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsMODEL PAPER 105 Marks
Question
A series L-C-R circuit is connected to an AC source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.
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Answer

Suppose a resistance R , inductance L and capacitance C in series. An alternating source of voltage $V = V _{ o }$ sin $\omega t$ is applied across it. Since all the components are connected in series, the current flowing through all is same.
Voltage across resistance R is $V _{ R }$, voltage across inductance L is $V _{ L }$ and voltage across capacitance C is $V _{ C }$. 
$V _{ R }$ and $\left( V _{ C }- V _{ L }\right)$ are mutually perpendicular and the phase difference between them is $90^{\circ}$.
Image

From the figure above, we have 
$V^2=V_R^2+\left(V_C-V_L\right)^2 \Rightarrow V=\sqrt{V_R^2+\left(V_C-V_L\right)^2} \ldots \ldots$ (i)
and $V_R=R i, V_C=X_C i$ and $V_L=V_L i \ldots$. (ii)
where $X _{ C }=\frac{1}{\omega C}=$ capacitance reactance and $X _{ L }=\omega L=$ inductive reactance
$\therefore \quad V=\sqrt{(R i)^2+\left(X_C i-X_L i\right)^2}$
$\therefore$ Impendance of circuit, $Z=\frac{V}{i}=\sqrt{R^2+\left(X_C-X_L\right)^2}$
i.e. $Z=\sqrt{R^2+\left(X_C-X_L\right)^2}=\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}$
The phase difference between current and voltage is given by, 
$\tan \phi=\frac{X_C-X_L}{R}$
Image

From the graph, we can see that with increase in frequency, current first increases and then decreases. At resonant frequency, current amplitude is maximum.

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