In a certain series LCR A.C. circuit, it is found that X L = 2 X C and phase difference between the current and voltage is 4 π . If now capacitance is made 1/ 4 th & the inductance and resistance are doubled then which are NOT CORRECT.

Initially, δ = 4 π ⇒ X 2 − X c = R = X c , Now, X c = 4 X c As X c 1 = X L 1 ⇒ δ 1 = 0

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PhysicsEMI/ACSeries AC CircuitsMedium2 minQB

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In a certain series LCR A.C. circuit, it is found that $X_{L} = 2 X_{C}$ and phase difference between the current and voltage is $\frac{π}{4}$ . If now capacitance is made $1/ 4^{th} &$ the inductance and resistance are doubled then which are NOT CORRECT.

Options:(select one or more)

Answer:

A, B, C

Solution:

Initially, $δ = \frac{π}{4} \Rightarrow X_{2} - X_{c} = R = X_{c}$ , Now, $X_{c} = 4 X_{c}$ As $X_{c}^{1} = X_{L}^{1} \Rightarrow δ^{1} = 0$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits

⏱ 2mℹ️ Source: QB

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