In a circuit, \(L, C\) and \(R\) are connected in series with an alternating voltage source of frequency \(f.\) The current leads the voltage by \(45^{\circ}\). The value of \(C\) will be:
1. | \(\dfrac{1}{2 \pi f \left( 2 \pi f L + R \right)}\) | 2. | \(\dfrac{1}{\pi f \left(2 \pi f L + R \right)}\) |
3. | \(\dfrac{1}{2 \pi f \left( 2 \pi f L - R \right)}\) | 4. | \(\dfrac{1}{\pi f \left(2 \pi f L - R \right)}\) |
In an LR-circuit, the inductive reactance is equal to the resistance R of the circuit. An e.m.f. applied to the circuit. The power consumed in the circuit is:
(1)
(2)
(3)
(4)
One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has a self-inductance of value: (f = 50 Hz)
(1) 0.052 H
(2) 2.42 H
(3) 16.2 mH
(4) 1.62 mH
In the circuit given below, what will be the reading of the voltmeter
(1) 300 V
(2) 900 V
(3) 200 V
(4) 400 V
In the circuit shown below, what will be the readings of the voltmeter and ammeter?
1. \(800~\text{V}, 2~\text{A}\)
2. \(300~\text{V}, 2~\text{A}\)
3. \(220~\text{V}, 2.2~\text{A}\)
4. \(100~\text{V}, 2~\text{A}\)
The diagram shows a capacitor C and a resistor R connected in series to an ac source. V1 and V2 are voltmeters and A is an ammeter:
Consider now the following statements
I. Readings in A and V2 are always in phase
II. Reading in V1 is ahead in phase with reading in V2
III. Readings in A and V1 are always in phase
Which of these statements is/are correct?
(1) I only
(2) II only
(3) I and II only
(4) II and III only
In the circuit shown in figure neglecting source resistance the voltmeter and ammeter reading will respectively, will be
(1) 0V, 3A
(2) 150V, 3A
(3) 150V, 6A
(4) 0V, 8A
In the circuit shown in the figure, the ac source gives a voltage Neglecting source resistance, the voltmeter and ammeter reading will be:
(1) 0V, 0.47A
(2) 1.68V, 0.47A
(3) 0V, 1.4 A
(4) 5.6V, 1.4 A
An ac source of angular frequency \(\omega\) is fed across a resistor \(r\) and a capacitor \(C\) in series. \(I\) is the current in the circuit. If the frequency of the source is changed to \(\frac{\omega}{3}\) (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency \(\omega\).
1. | \(\sqrt{\dfrac{3}{5}}\) | 2. | \(\sqrt{\dfrac{2}{5}}\) |
3. | \(\sqrt{\dfrac{1}{5}}\) | 4. | \(\sqrt{\dfrac{4}{5}}\) |
1. | \(\frac{\sqrt{5} R}{2} ,\tan^{- 1} \left(2\right)\) | 2. | \(\frac{\sqrt{5} R}{2} , \tan^{- 1} \left(\frac{1}{2}\right)\) |
3. | \(\sqrt{5} X_{C} ,\tan^{- 1} \left(2\right)\) | 4. | \(\sqrt{5} R , \tan^{- 1} \left(\frac{1}{2}\right)\) |