The power dissipated in an L-C-R series circuit connected to an AC source of emf E is:

1. \(\frac{\varepsilon^2R}{\Big[R^2+\Big(L\omega-\frac{1}{C\omega}\Big)^2\Big]}\)

2. \(\frac{\varepsilon^2\sqrt{R^2+\Big(L\omega-\frac{1}{C\omega}\Big)^2}}{R}~\)

3. \(\frac{\varepsilon^2\Big[R^2+\Big(L\omega-\frac{1}{C\omega}\Big)^2\Big]}{R}\)

4. \(\frac{\varepsilon^2R}{\sqrt{R^2+\Big(L\omega+\frac{1}{C\omega}\Big)^2}}~\)

In an AC circuit, the emf (e) and the current (I) at any instant are given respectively by 
e = E₀sin$\mathrm{\omega }$t 
I = I₀sin$\left(\mathrm{\omega t}-\mathrm{\varphi }\right)$ 
The average power in the circuit over one cycle of AC is:

1.  $\frac{{\mathrm{E}}_{\mathrm{o}}{\mathrm{I}}_{\mathrm{o}}}{2}$

2.  $\frac{{\mathrm{E}}_{\mathrm{o}}{\mathrm{I}}_{\mathrm{o}}}{2}$ $\sin$ $\mathrm{\varphi }$

3.  $\frac{{\mathrm{E}}_{\mathrm{o}}{\mathrm{I}}_{\mathrm{o}}}{2}$ $\cos$ $\mathrm{\varphi }$

4.  ${\mathrm{E}}_{\mathrm{o}}{\mathrm{I}}_{\mathrm{o}}$

If we change the value of \(R,\) then:
  

Voltage across each elements of a series LCR
circuit are given by V_L= 60V, V_C= 20V,V_R= 30V
Find out source voltage.
1. 50V 

2. 100V 

3. 150V 

4. 200V

In the given circuit, the reading of voltmeter V₁ and V₂ are 300 V each. The reading of the voltmeter V₃ and ammeter A are respectively:

1. 150 V, 2.2 A

2. 220 V, 2.2 A

3. 220 V, 2.0 A

4. 100 V, 2.0 A

An AC voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3 $\Omega$, the phase difference between the applied voltage and the current in the circuit is:

1.  $\frac{\mathrm{\pi }}{4}$

2.  $\frac{\mathrm{\pi }}{2}$

3.  zero

4.  $\frac{\mathrm{\pi }}{6}$

The current I, potential difference V_L across the inductor and potential difference V_C across the capacitor in the circuit as shown in the figure are best represented vectorially as:

1.                                      2.  

3.                                  4.