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Chapter 14: Problem 2102

In an LCR series ac circuit the voltage across each of the components \(\mathrm{L}, \mathrm{C}\) and \(\mathrm{R}\) is \(50 \mathrm{~V}\). The voltage across the LC combination will be (a) \(50 \mathrm{~V}\) (b) \(50 \sqrt{2} \mathrm{~V}\) (c) \(100 \mathrm{~V}\) (d) \(0 \mathrm{~V}\) (zero)

Short Answer

Expert verified
The voltage across the LC combination in an LCR series AC circuit with voltages of 50 V across L, C, and R components is 0 V (zero), as the voltages across the inductor (L) and capacitor (C) are out of phase by 180 degrees and cancel each other out when added using phasor addition. The correct answer is (d) \(0 \mathrm{~V}\) (zero).

Step by step solution

01

Identify the Voltages Across Components

The voltage across each of the components is given as 50 V. So, we have: \(V_L = 50 V\) \(V_C = 50 V\) Now, we will find the voltage across the LC combination (V_LC) using phasor addition.
02

Phasor Addition

In an LCR series AC circuit, the voltages across the inductor (L) and capacitor (C) are out of phase by 180 degrees. To find the resultant voltage across the LC combination, we need to add the voltages of L and C using phasor addition, considering their phase difference. Since they are out of phase by 180 degrees, the total voltage across LC is given by: \(V_{LC} = |V_L - V_C|\)
03

Calculating Voltage Across LC Combination

Now, we can plug in the values for V_L and V_C and calculate the voltage across the LC combination: \( V_{LC} = |50 V - 50 V|\) \(V_{LC} = |0 V|\) \(V_{LC} = 0 V\) So, the voltage across the LC combination will be 0V (zero). The correct answer is: (d) \(0 \mathrm{~V}\) (zero)

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Most popular questions from this chapter

In a region of uniform magnetic induction \(\mathrm{B}=10^{-2}\) tesla, a circular coil of radius \(30 \mathrm{~cm}\) and resistance \(\pi^{2}\) ohm is rotated about an axis which is perpendicular to the direction of \(\mathrm{B}\) and which forms a diameter of the coil. If the coil rotates at $200 \mathrm{rpm}$ the amplitude of the alternating current induced in the coil is, (a) \(4 \pi^{2} \mathrm{~mA}\) (b) \(30 \mathrm{~mA}\) (c) \(6 \mathrm{~mA}\) (d) \(200 \mathrm{~mA}\)

A wire of length \(2 \mathrm{~m}\) is moving at a speed \(2 \mathrm{~ms}^{-1}\) keep its length perpendicular to uniform magnetic field of \(0.5 \mathrm{~T}\). The resistance of circuit joined with this wire is \(6 \Omega\). The rate at which work is being done to keep the wire moving at constant speed is ........... (a) \((1 / 3) \infty\) (b) \((2 / 3) \omega\) (c) \((1 / 6) \propto\) (d) \(2 \mathrm{c}\)

A coil has an area of \(0.05 \mathrm{~m}^{2}\) \& it has 800 turns. It is placed perpendicularly in a magnetic field of strength, $4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{-2}\(. It is rotated through \)90^{\circ}\( in \)0.1 \mathrm{sec}$. The average emf induced in the coil is... (a) \(0.056 \mathrm{v}\) (b) \(0.046 \mathrm{v}\) (c) \(0.026 \mathrm{v}\) (d) \(0.016 \mathrm{v}\)

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