In an LR-circuit, the inductive reactance is equal to the resistance R of the circuit. An e.m.f. $E=E_0\cos \left(\omega t\right)$ applied to the circuit. The power consumed in the circuit is:

1. $\frac{E_0^2}{R}$

2. $\frac{E_0^2}{2R}$

3. $\frac{E_0^2}{4R}$

4. $\frac{E_0^2}{8R}$

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. \(100~\text{V}, 2~\text{A}\)

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 $V=20\cos \left(2000\text{\hspace{0.17em}}t\right).$ 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 \(\omega\).

In the adjoining ac circuit the voltmeter whose reading will be zero at resonance is

1. V₁

2. V₂

3. V₃

4. V₄

In the adjoining figure, the impedance of the circuit will be:

(1) 120 ohm

(2) 50 ohm

(3) 60 ohm

(4) 90 ohm