In an LCR series circuit, an inductor 30 mH and a resistor 1 are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which the current leads the voltage by is . Then the value of is

Question

In an LCR series circuit, an inductor 30 mH and a resistor 1 $Ω$ are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which the current leads the voltage by $45 °$ is $\frac{1}{x} \times 10^{- 3} F$ . Then the value of $x$ is _______.

Smart Achievers · Accepted Answer

(3)

$\tan ϕ = \frac{x_{C} - x_{L}}{R}$

$\tan 45 ° = \frac{x_{C} - x_{L}}{R}$

$x_{C} - x_{L} = R$

$\frac{1}{ω C} - ω L = R$

$\frac{1}{ω C} - 300 \times 0.03 = 1$

$\frac{1}{ω C} = 10$

$C = \frac{1}{10 ω} = \frac{1}{10 \times 300}$

$C = \frac{1}{3} \times 10^{- 3}$

$x = 3$

Solution

Q.
In an LCR series circuit, an inductor 30 mH and a resistor 1 $Ω$ are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which the current leads the voltage by $45 °$ is $\frac{1}{x} \times 10^{- 3} F$ . Then the value of $x$ is _______.

Ans.
(3)

$\tan ϕ = \frac{x_{C} - x_{L}}{R}$

$\tan 45 ° = \frac{x_{C} - x_{L}}{R}$

$x_{C} - x_{L} = R$

$\frac{1}{ω C} - ω L = R$

$\frac{1}{ω C} - 300 \times 0.03 = 1$

$\frac{1}{ω C} = 10$

$C = \frac{1}{10 ω} = \frac{1}{10 \times 300}$

$C = \frac{1}{3} \times 10^{- 3}$

$x = 3$

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Solution

Q. In an LCR series circuit, an inductor 30 mH and a resistor 1 Ω are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which the current leads the voltage by 45° is 1x×10-3 F. Then the value of x is _______.

Ans. (3) tanϕ=xC-xLR tan45°=xC-xLR xC-xL=R 1ωC-ωL=R 1ωC-300×0.03=1 1ωC=10 C=110ω=110×300 C=13×10-3 x=3

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Q.
In an LCR series circuit, an inductor 30 mH and a resistor 1 $Ω$ are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which the current leads the voltage by $45 °$ is $\frac{1}{x} \times 10^{- 3} F$ . Then the value of $x$ is _______.

Ans.
(3)

$\tan ϕ = \frac{x_{C} - x_{L}}{R}$

$\tan 45 ° = \frac{x_{C} - x_{L}}{R}$

$x_{C} - x_{L} = R$

$\frac{1}{ω C} - ω L = R$

$\frac{1}{ω C} - 300 \times 0.03 = 1$

$\frac{1}{ω C} = 10$

$C = \frac{1}{10 ω} = \frac{1}{10 \times 300}$

$C = \frac{1}{3} \times 10^{- 3}$

$x = 3$