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Series LCR circuit is considered one the most difficult concept.
92 Questions around this concept.
In a LCR circuit capacitance is changed from C to 2C . For the resonant frequency to remain unchanged, the inductance should be changed from L to
The equation of current in L-C-R circuit is
In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 k with C = 2
. The resonant frequency
is 200 rad/s. At resonance the voltage across L is
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A series AC circuit containing an inductor $(20 \mathrm{mH})$, a capacitor $(120 \mu \mathrm{~F})$ and a resistor $(60 \Omega)$ is driven by an AC source of $24 \mathrm{~V} / 50 \mathrm{~Hz}$. The energy dissipated in the circuit in 60 s is:
In LC circuit the inductance $L=40 \mathrm{mH}$ and capacitance $C=100 \mu F$. If a voltage $V(t)=10 \sin (314) t)$ is applied to the circuit , the current in the circuit is given as :
In a series LCR circuit, the inductive reactance (XL) is 10$\Omega$ and the capacitive reactance (XC) is 4$\Omega$. The resistance (R) in the circuit is 6$\Omega$.
The power factor of the circuit is:
In an LCR series a.c. circuit, the voltage across each of the components, L, C, and R is 50 V. The voltage across the LC combination will be
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The Phase difference in L-C circuit is
If represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless :
The unit of impedance is:
Series LCR circuit-
The Figure given above shows a circuit containing a capacitor, resistor and inductor connected in series through an alternating/sinusoidal voltage source.
As they are in series so the same amount of current will flow in all the three circuit components and for the voltage, vector sum of potential drop across each component would be equal to the applied voltage.
Let 'i' be the amount of current in the circuit at any time and $\mathrm{V}_{\mathrm{L}}, \mathrm{V}_{\mathrm{C}}$ and $\mathrm{V}_{\mathrm{R}}$ the potential drop across $\mathrm{L}, \mathrm{C}$ and R respectively then
$
\begin{aligned}
& \mathrm{v}_{\mathrm{R}}=\mathrm{iR} \rightarrow \text { Voltage is in phase with } \mathrm{i} \\
& \mathrm{v}_{\mathrm{L}=\mathrm{i} \omega \mathrm{~L}} \rightarrow \text { voltage is leading } \mathrm{i} \text { by } 90^{\circ} \\
& \mathrm{v}_{\mathrm{c}}=\mathrm{i} / \omega \mathrm{c} \rightarrow \text { voltage is lagging behind } \mathrm{i} \text { by } 90^{\circ}
\end{aligned}
$
By all these we can draw phasor diagram as shown below -
One thing should be noticed that we have assumed that VL is greater than VC which makes i lags behind V. If VC > VL then i lead V. So as per our assumption, their resultant will be (VL -VC). So, from the above phasor diagram V will represent resultant of vectors VR and (VL -VC). So the equation becomes -
$
\begin{aligned}
V & =\sqrt{V_R^2+\left(V_L-V_C\right)^2} \\
& =i \sqrt{R^2+\left(X_L-X_C\right)^2} \\
& =i \sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} \\
& =i Z
\end{aligned}
$
where,
$
Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}
$
Here, Z is called Impedance of this circuit.
Now come to the phase angle. The phase angle for this case is given as -
$
\tan \varphi=\frac{V_L-V_C}{V_R}=\frac{X_L-X_C}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}
$
Now from the equation of the phase angle three cases will arise. These three cases are -
(i) When,
$
\omega L>\frac{1}{\omega C}
$
then, $\tan \varphi$ is positive i.e. $\varphi$ is positive and voltage leads the current i .
(ii) When $\omega L<\frac{1}{\omega C}$
then, $\tan \varphi$ is negative i.e. $\varphi$ is negative and voltage lags behind the current $i$.
(iii) When $\omega L=\frac{1}{\omega C}$,
then tanφ is zero i.e. φ is zero and voltage and current are in phase. This is called electrical resonance.
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