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 an \(LCR\) circuit, the potential difference between the terminals of the inductance is \(60\) V, between the terminals of the capacitor is \(30\) V and that between the terminals of the resistance is \(40\) V. The supply voltage will be equal to:
1. \(50\) V

2. \(70\) V

3. \(130\) V

4. \(10\) V

In a series \(RLC\) circuit, potential differences across \(R,L\) and \(C\) are \(30\) V, \(60\) V and \(100\) V respectively, as shown in the figure. The emf of the source (in volts) will be:

1. \(190\)

2. \(70\)

3. \(50\)

4. \(40\)

In an \(LCR\) series network, \(V_L = 40~\text{V}, V_C = 20~\text{V}~\text{and}~V_R = 15~\text{V}.\) The supply voltage will be:
1. \(25~\text{V}\)
2. \(75~\text{V}\)
3. \(35~\text{V}\)
4. zero

A series AC circuit has a resistance of \(4~\Omega\) and an inductor of reactance \(3~\Omega\). The impedance of the circuit is \(z_1\). Now when a capacitor of reactance \(6~\Omega\) is connected in series with the above combination, the impedance becomes \(z_2\). Then \(\frac{z_1}{z_2}\) will be:
1. \(1:1\)
2. \(5:4\)
3. \(4:5\)
4. \(2:1\)

In the circuit shown in the figure, neglecting source resistance, the voltmeter and ammeter reading respectively will be:

1. \(0~\text{V}, 3~\text{A}\)
2. \(150~\text{V}, 3~\text{A}\)
3. \(150~\text{V}, 6~\text{A}\)
4. \(0~\text{V}, 8~\text{A}\)

In a series \(LCR\) circuit, resistance \(R=10~\Omega\) and the impedance \(Z=20~\Omega\). The phase difference between the current and the voltage will be:
1. \(30^{\circ}\)
2. \(45^{\circ}\)
3. \(60^{\circ}\)
4. \(90^{\circ}\)