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Chapter 14: Problem 2088

An LCR series circuit with \(\mathrm{R}=100 \Omega\) is connected to a $200 \mathrm{~V}\(, \)50 \mathrm{~Hz}$ a.c source when only the capacitance is removed the current lies the voltage by \(60^{\circ}\) when only the inductance is removed, the current leads the voltage by \(60^{\circ}\). The current in the circuit is, (a) \(2 \mathrm{~A}\) (b) \(1 \mathrm{~A}\) (c) \((\sqrt{3} / 2) \mathrm{A}\) (d) \((2 / \sqrt{3}) \mathrm{A}\)

Short Answer

Expert verified
The current in the LCR series circuit is (a) 2 A.

Step by step solution

01

Write the given information

We are given: - Resistance, R = 100 Ω - Voltage, V = 200 V - Frequency, f = 50 Hz - Phase shifts when only capacitance or inductance is removed
02

Analyze the two different cases

Case 1 (Capacitance is removed): The current lags the voltage by 60° Case 2 (Inductance is removed): The current leads the voltage by 60°
03

Find reactances for each case

For each case, we need to find the reactance of the component that is left in the circuit (either inductive or capacitive) using the given phase shift information. Case 1: \(tan(\theta_{1}) = \frac{X_{L}}{R}\) Case 2: \(tan(\theta_{2}) = \frac{X_{C}}{R}\) Now we substitute the given data, \(\theta_{1} = 60°\) and \(\theta_{2} = -60°\), and find the respective inductive and capacitive reactances, \(X_{L}\) and \(X_{C}\).
04

Calculate the inductive reactance

We know that \(\tan(60°) = \sqrt{3}\), so: \( X_{L} = R \tan(60°) = 100\Omega \times \sqrt{3} = 100\sqrt{3} \Omega \)
05

Calculate the capacitive reactance

We know that \(\tan(-60°) = -\sqrt{3}\), so: \( X_{C} = - R \tan(-60°) = 100\Omega \times \sqrt{3} = 100\sqrt{3} \Omega \)
06

Calculate the net reactance (impedance) in the LCR circuit

We can calculate the net reactance (impedance) in the LCR circuit using the formula: \( Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}} = \sqrt{100^2 + (100\sqrt{3} - 100\sqrt{3})^2} = 100 \Omega \)
07

Calculate the current in the LCR circuit

We can calculate the current in the LCR circuit using Ohm's Law: \( I = \frac{V}{Z} = \frac{200 V}{100 \Omega} = 2 A \) Therefore, the correct answer is: (a) 2 A

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Most popular questions from this chapter

A coil of inductance \(300 \mathrm{mH}\) and resistance \(2 \Omega\) is connected to a source of voltage \(2 \mathrm{~V}\). The current reaches half of its steady state value in \(\ldots \ldots\) (a) \(0.15 \mathrm{sec}\) (b) \(0.3 \mathrm{sec}\) (c) \(0.05 \mathrm{sec}\) (d) \(0.1 \mathrm{sec}\)

A ring of radius \(\mathrm{r}\) is rotating about its diameter with angular velocity w in a perpendicular magnetic field \(\mathrm{B}^{-}\) It has 20 turns. The emf induced is (a) \(20 \mathrm{~B} \pi \mathrm{r}^{2} \sin \mathrm{wt}\) (b) \(20 \mathrm{~B} \pi \mathrm{r}^{2} \mathrm{cos} \mathrm{wt}\) (c) \(10 \sqrt{2} \mathrm{~B} \pi \mathrm{r}^{2}\) (d) \(20 \mathrm{~B} \pi \mathrm{r}^{2} \mathrm{w} \sin \mathrm{wt}\)

A magnetic field \(2 \times 10^{-2} \mathrm{~T}\) acts at right angles to a coil of area \(200 \mathrm{~cm}^{2}\) with 25 turns. The average emf induced in the coil is \(0.1 \mathrm{v}\) when it removes from the field in time t. The value of \(\mathrm{t}\) is (a) \(0.1 \mathrm{sec}\) (b) \(1 \mathrm{sec}\) (c) \(0.01 \mathrm{sec}\) (d) \(20 \mathrm{sec}\)

The time constant of a LR circuit is \(80 \mathrm{~ms}\), The circuit is connected at \(t=0\) and the steady state current is found to be $4 \mathrm{~A}\(. Find the current at \)20 \mathrm{~ms}$. (a) \(0.98 \mathrm{~A}\) (b) \(1 \mathrm{~A}\) (c) \(0.44 \mathrm{~A}\) (d) \(0.88 \mathrm{~A}\)

In an ac circuit the emf (e) and the current (i) at any instant core given respectively by \(\mathrm{e}=\mathrm{E}_{\mathrm{O}}\) sin $\operatorname{ct}, \mathrm{I}=\mathrm{I}_{\mathrm{O}} \sin (\cot -\Phi)$. The average power in the circuit over one cycle of ac is. (a) \(\left[\left\\{E_{O} I_{O}\right\\} / 2\right] \cos \Phi\) (b) \(\mathrm{E}_{\mathrm{O}} \mathrm{I}_{\mathrm{O}}\) (c) \(\left[\left\\{E_{Q} I_{Q}\right\\} / 2\right]\) (d) \(\left[\left\\{E_{O} I_{O}\right\\} / 2\right] \sin \Phi\)
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