A transformer is used to light a 100 W and 110 V lamp using a 220 V main supply. If the supply current is 0.5 A, then the efficiency of the transformer is

• 100 %

• 99 %

• 90.1 %

• 87.7 %

The power factor of L-C-R circuit at resonance is

• 0

• $\frac{1}{2}$

• $\frac{1}{\sqrt{2}}$

• 1

If E = 100 sin (100 t) volt and $\mathrm{I} = 100 \sin \left(100 \mathrm{t} + \frac{\mathrm{\pi }}{3}\right) \mathrm{mA}$ are the instantaneous values of voltage and current, then the rms values of voltage and current are respectively

• 70.7 V, 70.7 mA

• 70.7 V, 70.7 A

• 141.4 V, 141.4 mA

• 141.4 V, 141.4 A

The core of a transformer is laminated to reduce

• flux leakage

• output power

• eddy current

• copper loss

If E₀ is the peak emf, I_O is the peak current and is the phase diference between them, then the average power dissipation in the circuit is

• $\frac{1}{2} {\mathrm{E}}_0{\mathrm{I}}_0$

• $\frac{{\mathrm{E}}_0{\mathrm{I}}_0}{\sqrt{2}}$

• $\frac{1}{2} {\mathrm{E}}_0{\mathrm{I}}_0 \sin \mathrm{\varphi }$

• $\frac{1}{2} {\mathrm{E}}_0{\mathrm{I}}_0 \cos \mathrm{\varphi }$

An L-C-R series AC circuit is at resonance with 10 V each across L, C and R. If the resistance is halved, the respective voltages across L, C and R are

• 10 V, 10 V and 5 V

• 10 V, 10 V and 10 V

• 20 V, 20 V and 5 V

• 20 V,20 V and 1O V

A transformer is used to light a 100 W and 110 V lamp from a 220 V main supply. If the main current is 0.5 A, then the efficiency of the transformer is nearly

• 89 %

• 100 %

• 91 %

• 85 %

An L-C-R series circuit is at resonance. Then

• the phase difference between current and voltage is 90°

• the phase difference between current and voltage is 45°

• its impedance is purely resistive

• its impedance is zero

The impedance of a R-C circuit is Z₁ for frequency f and Z₂ for frequency 2f. Then Z₁/Z₂ is

• between 1 and 2

• 2

• 2 between $\frac{1}{2}$ and 1

• $\frac{1}{2}$

In an L-C-R series AC circuit the voltage across L, C and R is 10 V each. If the inductor is short circuited, the voltage across the capacitor would become

• 10 V

• $\frac{20}{\sqrt{2}} \mathrm{V}$

• $20\sqrt{2} \mathrm{V}$

• $\frac{10}{\sqrt{2}} \mathrm{V}$