Alternating Current

An alternating voltage, $E=200\sqrt{2}\sin \left(100\mathrm{t}\right)$  is connected to a $1 \mu \mathrm{F}$ capacitor through an A.C. ammeter. The reading of the ammeter shall be

A.C. voltmeter is connected to a source of peak voltage ${\mathrm{e}}_0=141.4 \mathrm{V}$ , then it will read

An A.C. generator produced an output voltage $e=170\sin 377\mathrm{t} \mathrm{V}$ , where $t$ is in seconds. The frequency of A.C. voltage is
 

An A.C. voltage$V=5\cos \left(1000t\right)\mathrm{V}$ is applied to a L-R series circuit of inductance $3 \mathrm{mH}$ and resistance $4 \mathrm{\Omega }$. The value of maximum current in the circuit is Ampere. 

A generator produces a voltage that is given by ${\mathrm{e}}_{.}=240\sin 120\mathrm{t}$,  where $t$ is in seconds. The frequency and r.m.s voltage are 

In an A.C. circuit, peak value of voltage is $423 \mathrm{V}$. Its effective voltage is 

The number of turns in the coil of an A.C. generator are $100$ and its cross-sectional area is $2.5 {\mathrm{m}}^2$ . The coil is revolving in a uniform magnetic field of strength $0.3 \mathrm{T}$ with the uniform angular velocity of $60 \mathrm{rad} {\mathrm{s}}^{-1}$. The value of maximum voltage produced in $\mathrm{kV}$ is

A multimeter reads a voltage of a certain A.C. source as$100 \mathrm{V}$. What is the peak value of voltage of A.C. source?
 

For an A.C. given by $I=50\cos \left(100t+45°\right) \mathrm{A}$ . The value of  ${\mathrm{I}}_{\mathrm{rms}}=$____$\mathrm{A}.$
 

If the instantaneous current is given by, $I=4\cos (\omega t+\mathrm{\varphi }) \mathrm{A}$ , then the r.m.s. value of the current is
 

E.m.f. is given by, $e=200\sin 50\mathrm{t}$. The r.m.s. value of current (in Ampere) in a circuit of resistance $50 \mathrm{\Omega }$ is
 

The ratio of peak value and r.m.s value of an alternating current is  
 

The frequency of A.C. mains in India is 
 

A capacitor of capacitance $10 \mathrm{\mu F}$ is connected to an $\mathrm{AC}$ source and an $\mathrm{AC}$ Ammeter. If the source voltage varies as $\mathrm{V}=50\sqrt{2}\sin 100\mathrm{t}$ the reading of the ammeter 

An alternating voltage $\mathrm{e}=200\sqrt{2}\sin \left(100\mathrm{t}\right) \mathrm{V}$  is connected to$1 \mathrm{\mu F}$ the capacitor through a.c. ammeter. The reading of the ammeter is 
 

The RMS value of current in a $50 \mathrm{Hz} \mathrm{AC}$ circuit is $6 \mathrm{A}$. The average value of $\mathrm{AC}$ current over a cycle is

If ${\epsilon }_o$  represents the peak value of the voltage in an A.C. circuit, the r.m.s. value of the voltage will be 
 

Alternating current cannot be measured by D.C. ammeter because

The r.m.s. value of induced voltage in a coil of $50$ turns each of area $30 {\mathrm{cm}}^2$ rotating at $1000 \mathrm{r}.\mathrm{p}.\mathrm{m}.$ about an axis at right angles to magnetic field is $\left(B=5\times {10}^{-4} \mathrm{Wb} {\mathrm{m}}^{-2}\right)$,
 

The r.m.s value of an A.C. of $50 \mathrm{Hz}$  is $10 \mathrm{A}$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be,