In the transformer shown in the figure, the ratio of the number of turns of the primary to the secondary is \(\dfrac{N_1}{N_2}= \dfrac{1}{50}.\) If a voltage source of \(10~\text V\) is connected across the primary, then the induced current through the load of \(10~\text{k}\Omega\) in the secondary is:
             
1. \(\dfrac{1}{20}~\text{A}\)
2. zero
3. \(\dfrac{1}{10}~\text{A}\)
4. \(\dfrac{1}{5}~\text{A}\)

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. \(15~\text A\) 

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:
    

An AC voltage source is connected to a series \(LCR\) circuit. When \(L\) is removed from the circuit, the phase difference between current and voltage is \(\dfrac{\pi}{3}\). If \(C\) is instead removed from the circuit, the phase difference is again \(\dfrac{\pi}{3}\) between current and voltage. The power factor of the circuit is:
1. \(0.5\)
2. \(1.0\)
3. \(-1.0\)
4. zero

The circuit is in a steady state when the key is at position \(1\). If the switch is changed from position \(1\) to position \(2\), then the steady current in the circuit will be:
  

In a box \(Z\) of unknown elements (\(L\) or \(R\) or any other combination), an ac voltage \(E = E_0 \sin(\omega t + \phi)\) $$is applied and the current in the circuit is found to be \(I = I_0 \sin\left(\omega t + \phi +\frac{\pi}{4}\right)\). The unknown elements in the box could be: