Applied voltage in any ac circuit is 220 V . If $R =8 \Omega, X _{ L }= X _{ C }=6 \Omega$, then find the value of the following :(a) root mean square value of the voltage, (b) impedance of the circuit.
Answer
(a) $ V_{rms}=220 V $ (b) Impedance of circuit $\quad Z =\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2}$ Now, $X_L=X_C$ then $Z=R$ Impedance of circuit $Z=8 \Omega$.
Write the definition of root mean square value of alternating current.
Answer
The square root of the mean square of the instantaneous value of alternating curent is called the root mean square value $I_{ rms }$ of the current. $I ^2= I _0^2 \sin ^2 \omega t$ $I _{ rms }=\frac{ I _0}{\sqrt{2}}$ or $I _{ rms }=0.707 I _0$ Root mean square value of ac current is obtained by multiplying its peak value by $\frac{1}{\sqrt{2}}$ or 0.707
Calculate power loss in an ac circuit in which magnitudes of voltage and current are as follows :$V =3000 \sin \left(\omega t+\frac{\pi}{2}\right)$ and $I =5 \sin \omega t$
Answer
Here $\phi=\frac{\pi}{2}$ $P = E _{ rms } I _{ rms } \cos \phi$ $\therefore \quad P=0$ $\because \cos \frac{\pi}{2}=0$
What is the principle on which a transformer works? Can it be used in de circuit?
Answer
It works on the principle of electromagnetic induction. It cannot work in de circuit because direct current cannot produce variable magnetic field in core of the transformer.
A capacitor and a bulb are joined in series with an ac voltage source. If frequency of alternating emf is increased then what is the effect on the circuit?
Answer
On increasing the frequency, reactance of capacitor $\frac{1}{ C \omega}$ decreases due to which current in the circuit increases and brightness of bulb increases.
What is the meaning of instantaneous power and average power in an ac circuit?
Answer
The magnitude of power at any instant in an ac circuit is known as instantaneous power and average value of instantaneous powers in one complete cycle is known as average power.
Resistance of a coil is 30 and its reactance is 42. What is power factor of the coil?
Answer
Power factor $=\frac{R}{Z}$ But $Z =\sqrt{( R )^2+\text { Reactance }^2}$ $=\sqrt{(3)^2+(4)^2}=5$ $\therefore \quad$ Power factor $=\frac{R}{Z}=\frac{3}{5}$ Ans.
Voltage across the ends of inductor, capacitor and resistor in a series LCR circuit is 20 V, 20 V and 40 V respectively. What is the phase difeerence between the current and applied voltage in the circuit?
Answer
$\tan \phi=\left(\frac{ V _{ C }- V _{ L }}{ V }\right)=\left(\frac{20-20}{40}\right)=0$ $\phi=\tan ^{-1}(0)=0$
If an ac circuit, $R=100 \Omega, X_L=400 \Omega$ and $X _{ C }=400 \Omega$, then what is the total impedance of the circuit?
Answer
Impedance $Z=\left( R ^2+\left( X _{ L }- X _{ C }\right)^2\right)^{1 / 2}$ When $X_L=X_C $ then $Z =\left( R ^2+0\right)^{1 / 2}= R$ $\therefore \quad Z=100 \Omega$
When an electrical component $X$ is joined with any ac circuit, then current in it is ahead of phase of voltage by $\frac{\pi}{2}$ radian. Identify X and write an expression for its reactance.
Answer
Component X is a capacitor and its reactance
$ =\frac{1}{\omega C}=\frac{1}{2 \pi / C}
$
Is it possible in a LCR series circuit that the value of applied potential on any component of circuit is more than the peak value of applied alternating potential?
The instantaneous magnitudes of current and voltage in an ac circuit are given as $I =10 \sin 314 t A$ and $V =50 \sin \left(314 t +\frac{\pi}{2}\right) \quad V$ respectively. What is the dissipation of average power in the circuit?
Answer
Phase difference between voltage and current $=\frac{\pi}{2}$ $\therefore \quad \cos \phi=\cos \frac{\pi}{2}=0$ $\therefore \quad$ Average power $\quad \begin{aligned} \overline{\mathrm{P}} =\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi \\ =\mathrm{V}_{\mathrm{rms}} \times \mathrm{I}_{\mathrm{rms}} \times 0=0\end{aligned}$
No, because alternating current changes periodically whereas for the process of electrolysis, the magnitude nad direction of current does not change with time.
In an ac circuit, instantaneous current and voltage are $I = I _0 \sin 300 t A$ and $V =200 \sin 300 t$ respectively. What is average power dissipated in the circuit?
Answer
It is clear from the given expressions that phase difference $\phi$ between current and voltage is $\phi=0^{\circ}$. $\overline{ P }= V _{ rms } \cdot I _{ rms } \cos \phi$ $=\frac{V_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \times \cos 0^{\circ}$ $=\frac{200}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times 1$ $=1000 W$
What is the average power given by an ac source to a capacitor?
Answer
Phase difference between voltage and current in a capacitor $=\phi=90^{\circ}$ Average power $\overline{ P }= E _{ mxx } I _{ rms } \cos \phi=0$$\because \cos 90^{\circ}=0$
An ac current $I = I _0 \sin \omega t$ produces some heat H in any resistor R in time $T =\frac{2 \pi}{\omega}$. Write the magnitude of dc current which will produce same heat in same resistor in same time.
Answer
Required dc current $=$root mean square value of alternating current $ I_{rms}=\frac{I_0}{\sqrt{2}} $ where $I_0$ is peak value of alternating current.
Expression $E = E _0 \sin \omega t$ represents alternating electromotive force. Find its frequency, magnitude and time period.
Answer
Angular frequency $\omega=2 \pi f,(f)=\frac{\omega}{2 \pi}$ Magnitude $=$ Peak value of voltage $= E _0$ Time period, $T =\frac{1}{f}=\frac{2 \pi}{\omega}$
When a transformer is known as an ideal transformer?
Answer
In an ideal transformer, there is no loss in energy on transferring the energy from primary coil to secondary coil and then the value of power is same in both the coils. Efficiency of such transformer should be $100 \%$.