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  4. An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance ' R ' are connected in series to an ac source of potential difference ' V ' volts as shown in figure. Potential difference across L , C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10 √2 A. The impedance of the circuit is :

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Q. An inductor of inductance $L$, a capacitor of capacitance $C$ and a resistor of resistance ' $R$ ' are connected in series to an ac source of potential difference ' $V$ ' volts as shown in figure. Potential difference across $L , C$ and $R$ is $40 V$, $10\, V$ and $40 V$, respectively. The amplitude of current flowing through LCR series circuit is $10 \sqrt{2} A$. The impedance of the circuit is :Physics Question Image

NEETNEET 2021Alternating Current

Solution:

Given:
$V_{L}=40$ volt; $V_{R}=40$ volt; $V_{C}=10$ volt Now,
$V _{ RMS }=\sqrt{ V _{ R }^{2}+\left( V _{ L }- V _{ C }\right)^{2}}$
$=\sqrt{(40)^{2}+(40-10)^{2}}=50 V$
$I _{ RMS }=\frac{ I _{ O }}{\sqrt{2}}=\frac{10 \sqrt{2}}{\sqrt{2}}=10 A$
$ \because V _{ RMS }= I _{ RMS } \times Z$
$\therefore Z=\frac{V_{ RMS }}{ l _{ RMS }}=\frac{50}{10}=5 \Omega$