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Byju's Answer

Standard XII

Physics

RLC Circuit

The current f...

Question

The current flowing through the resistor in a series LCR a.c. circuit, is $I = ε / R .$
Now the inductor and capacitor are connected in parallel and joined in series with the resistor as shown in figure. The current in the circuit is now. (Symbols have their usual meaning)

equal to I

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more than I

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less than I

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zero

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Solution

The correct option is D zero
In figure (1) $Z = R$
$X_{L} = X_{C}$
If $I_{1} = \frac{V_{0}}{X_{C}} s i n (ω t + π / 2)$
then $I_{2} = \frac{V_{0}}{X_{L}} s i n (ω t - π / 2)$
and Total current in circuit two $I = I_{1} + I_{2} = 0$

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The current flowing through the resistor in a series LCR a.c. circuit, is I=ε/R.Now the inductor and capacitor are connected in parallel and joined in series with the resistor as shown in figure. The current in the circuit is now. (Symbols have their usual meaning)

The correct option is D zeroIn figure (1) Z=R XL=XCIf I1=V0XCsin(ωt+π/2)then I2=V0XLsin(ωt−π/2)and Total current in circuit two I=I1+I2=0

The current flowing through the resistor in a series LCR a.c. circuit, is $I = ε / R .$
Now the inductor and capacitor are connected in parallel and joined in series with the resistor as shown in figure. The current in the circuit is now. (Symbols have their usual meaning)

The correct option is D zero
In figure (1) $Z = R$
$X_{L} = X_{C}$
If $I_{1} = \frac{V_{0}}{X_{C}} s i n (ω t + π / 2)$
then $I_{2} = \frac{V_{0}}{X_{L}} s i n (ω t - π / 2)$
and Total current in circuit two $I = I_{1} + I_{2} = 0$