In the circuit shown, the AC source has a voltage
\(V= 20\cos(\omega t)\) volts with \(\omega = 2000\) rad/s. The amplitude of the current will be nearest to:

1. \(2\) A
2. \(3.3\) A
3. \(\frac{2}{\sqrt{5}}\) A
4. \(\sqrt{5}\) A

The AC source in the circuit shown in the figure produces a voltage \(V = 20\cos(2000t)\) volts. Neglecting source resistance, the voltmeter and ammeter readings will be (approximately):

An AC ammeter is used to measure the current in a circuit. When a given direct current passes through the circuit, the AC ammeter reads \(6~\text A.\) When another alternating current passes through the circuit, the AC ammeter reads \(8~\text A.\) Then the reading of this ammeter if DC and AC flow through the circuit simultaneously is:
1. \(10 \sqrt{2}~\text A\) 
2. \(14~\text A\) 
3. \(10~\text A\) 
4. \(15~\text A\) 

If \(q\) is the capacitor's charge and \(i\) is the current at time \(t\), the voltage \(V\) will be:

A direct current of \(5~ A\) is superimposed on an alternating current \(I=10sin ~\omega t\) flowing through a wire. The effective value of the resulting current will be:

An ideal resistance \(R,\) ideal inductance \(L,\) ideal capacitance \(C,\) and AC voltmeters ${\mathrm{}}_{}$\(V_1, V_2, V_3~\text{and}~V_4 \)are connected to an AC source as shown. At resonance:
    

A transistor-oscillator using a resonant circuit with an inductance \(L\) (of negligible resistance) and a capacitance \(C\) has a frequency \(f\). If \(L\) is doubled and \(C\) is changed to \(4C\), the frequency will be:
1. \(\frac{f}{4}\)
2. \(8f\)
3. \(\frac{f}{2\sqrt{2}}\)
4. \(\frac{f}{2}\)