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Physics 5 Marks Questions

Alternating Current · CBSE STD 12 Science (CBSE - English Medium). 17 questions.

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Question 15 Marks
(a) Prove that the mean power supplied to the inductor in a complete cycle is zero. (b) If in an LCR alternatin current circuit $R =$ $24 \Omega, X _{ L }=110 \Omega$, and $X _{ C }=110 \Omega$, then find the impednce of the current.
Answer
(a) Average power : Instantancous power of a circuit with pure inductance
$P=E \times I$
$= E _0 \sin \omega t \times I _0 \sin \left(\omega t-\frac{\pi}{2}\right)$
$= E _0 \sin \omega t \times\left[- I _0 \cos \omega t\right]$
$\because \sin \left(\theta-90^{\circ}\right)$ is $=\cos \theta$
$=- E _0 I _0 \sin \omega t \cos \omega t$
or $P =\frac{- E _0 I _0}{2} \times 2 \sin \omega t \cos \omega t$
$=-\frac{ E _0 I _0}{2} \times \sin 2 \omega t$
$[\because \sin 2 A=2 \sin A \cos A ]$
The average value of $\sin \omega t$ in a complete cycle is zero, that is
$\overline{\sin 2 \omega t}=\frac{\int_0^{ T } 2 \sin \omega t d t}{\int_0^{ T } d t}=0$
Therefore the average power of the circuit in one complete cycle
$P=\frac{E_0 I_0}{2} \times 0=0$
Therefore, the average power supplied to the inductance in a complete cycle is zero.
(b) Given, that $R =24 \Omega, X _{ L }=110 \Omega, X _{ C }=110 \Omega$
$Z =\sqrt{ R ^2+\left( X _{ L }^2- X _{ C }^2\right)^2}$
$X _{ L }= X _{ C }=100 \Omega$
$Z =\sqrt{(24)^2+(0)^2}$
$Z=24$
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Question 25 Marks
(a) Prove that the peak value ( $I_m$ ) of an alternating current is $\sqrt{2}$ times its root mean square ( rms ) value.(b) If alternating current $I=4 sin \omega t$ of and voltage $V =200 \sin (\omega t+\pi / 3)$, calculate the dissipated average power in the circuit.
Answer
(a) The root mean square (r.m.s.) value of alternating current is equal to that value of direct current. Which exhibits the same heating effect as alternating current. We represent this by $I _{ mms }$ or $I _{ eff }$
As a mathematical definition, the square root of the mean square of the instantaneous alternating current for a complete cycle is called the root mean square of the alternating current (r.m.s.)
Equation of instantaneous alternating current $ I=I_0 \sin \omega t $
On squaring the instantaneous value of alternating current, $ I^2=I_0{ }^2 \sin ^2 \omega t $
Mean value of $I ^2$ will be for one periodic time.
$I ^2=$ Mean value of $I ^2$ will be for one periodic time/ Time period
$\overline{ I ^2}=\frac{\int_0^{ T } I ^2 d t}{\int_0^{ T } d t}=\frac{1}{T} \int_0^{ T } I ^2 \sin ^2 \omega t d t$
$\overline{ I ^2}=\frac{ I _0^2}{T} \int_0^{ T } \sin ^2 \omega t d t$
$=\frac{ I _0^2}{T} \int_0^{ T }\left[\frac{1-\cos 2 \omega t}{2}\right] d t$
$\because \quad \sin ^2 \omega t=\frac{1-\cos 2 \omega t}{2}$
$=\frac{ I _0^2}{2 T} \int_0^{ T }[1-\cos 2 \omega t] d t$
$=\frac{ I _0^2}{2 T}\left[\int_0^{ T } d t-\int_0^{ T } \cos 2 \omega t d t\right]$
$=\frac{ I _0^2}{2 T}\left[[t]_0^{ T }-\left(\frac{\sin 2 \omega t}{2 \omega}\right)_0^{ T }\right]$
$=\frac{I_0^2}{2 T}\left[(T-0)-\frac{1}{2 \omega}[\sin 2 \omega T-\sin 0]\right]$
$[\because \omega t=2 \pi]$
$=\frac{I_0^2}{2 T}\left[T-\frac{1}{2 \omega}\left(\frac{\sin 2 \omega \times 2 \omega}{\omega}-\sin 0\right)\right]$
$=\frac{I_0^2}{2 T}\left(T-\frac{1}{2 \omega}(\sin 4 \pi-\sin 0)\right)$
$=\frac{ I _0^2}{2 T}\left( T -\frac{1}{2 \omega}(0-0)\right)$
$\overline{ I ^2}=\frac{ I _0^2}{2 T} \times T =\frac{ I _0^2}{2}\left[\because \sin 4 \pi=\sin 0^{\circ}=0\right]$
Taking square root
$\overline{I^2}=\frac{I_0}{\sqrt{2}}$
$I _{ rms }=0.707 I _0$ $\left[\because \frac{1}{\sqrt{2}}=0.707\right]$
$I _{ rms }=70.7 \%$ of $I _0$
It is clear from equation (4) that the root mean square value of alternating current is $70.7 \%$ of the peak value ( $I _0$ ) of the current. This can be expressed through the following graph :
Image
(b) $\quad V_{\text {rms }}=\frac{V_0}{\sqrt{2}}=0.707 E _0$
$=0.707 \times 200$
$=141.4$ volt
(Here $V _0=200$ , $I _0=4$)
$I_{ rms }=\frac{ I _0}{\sqrt{2}}=0.707 I _0$
$=0.707 \times 4$ $=2.878 A$
$\phi=\pi / 2=60^{\circ}$
$\cos \theta=\cos 60^{\circ}=0.5$
Average power
$\overline{ P }=\frac{ V _0}{\sqrt{2}} \times \frac{ I _0}{\sqrt{2}} \cos \phi$
$=\frac{V_0 I_0}{2} \cos \phi$
$=\frac{200 \times 4}{2} \times \frac{1}{2}$
$P =200 W$
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Question 35 Marks
Match the following
Image
Answer
(i) $\frac{1}{\sqrt{ LC }}$
(ii) $\frac{\omega_0 L}{ R }$
(iii)$VI \cos \phi$
(iv)$\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2}$
(v) $\frac{1}{2} LI ^2$
(vi) $\frac{- E }{\left(\frac{ dI }{ dt }\right)}$
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Question 45 Marks
(a) Obtain an expression for current flowing through an ideal inductor joined with an ac source of voltage $v=v_0 \sin \omega t$ by using phasor diagram. Draw a graph between (i) voltage applied and (ii) current as a function of (wt.).
(b) Obtain an expression for average power dissipation in any series LCR circuit.
Answer
SELF
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Question 55 Marks
Draw a phasor diagram for series LCR circuit joined with ac voltage source and obtain an expression for impedance of the circuit.
Answer
SELF
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Question 65 Marks
What is a transformer? It is of how many types? Explain the principle, construction and working of any transformer.
Answer
SELF
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Question 75 Marks
Define power and explain the following points (a) Instantaneous power (b) average power (c) power factor
Answer
SELF
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Question 105 Marks
Exlplain ac voltage applied on a series LCR circuit. Explain its solution by phasor diagram.
Answer
SELF
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Question 125 Marks
Explain ac voltage applied on inductor and explain that when only pure inductor is joined in a circuit, the current lags emf by a phase angle of $\frac{\pi}{2}$.
Answer
SELF
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Question 135 Marks
Write advantages and disadvantages of ac over dc and explain applied ac voltage over resistor.
Answer
SELF
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Question 145 Marks
Obtain expression for root mean square value of alternating emf. Draw graph between $E$ and t and $E ^2$ and $t$ and prove that average rate of heat produced for alternating current in one complete cycle is $H = I _{\text {max }}^2 R$.
Answer
SELF
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Question 165 Marks
Explain direct current direct emf and draw their graphs. Define alternating current and explain in detail.
Answer
SELF
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Question 175 Marks
Obtain expressions for average current and root mean square current for a complete cycle in ac circuit.###Obtain root mean square value of current for one complete cycle of current $I=I_0 \sin \omega t$. Draw graph between instantaneous value and root mean square value of current for two complete cycles.###Obtain rms value of alternating current for a complete cycle of sinusoidal current.$I = I _0 \sin \omega t .$
Answer
SELF
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