2 Marks Questions

Question 12 Marks

Sunita and her friends went to an exhibition. The guard standing there asked them to pass through a metal detector. Sunita's friends were first afraid but when Sunita told them the reason and explain its working.
Answer the following questions :
(a) State the principle on which metal detector works.
(b) If a person passing through metal detector has some metallic object then why does it produces sound?
(c) Tell two qualities of Sunita which she showed while explaining the reason for passing through it.

Answer

(a) Metal detector works on the principle of resonance in ac circuit.
(b) Impedance of the circuit changes and as a result, resultant current in the circuit changes and this is the reason that the metal detector produces sound.
(c) (i) Knowledge, (ii) scientific temper.

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Question 22 Marks

Explain the root mean square value of alternating current.

Answer

Root mean square value : The square root of the mean square of the instantaneous value of alternating curent is called the root mean square value $I_{\text {rms }}$ of the current. Root mean square value of alternating current is equal to the magnitude of direct current which shows the same heating effect as shown by alternating current. It is denoted by $I _{ rms }$.
$I _{ rms }=\frac{ I _0}{\sqrt{2}}$
$I _{ rms }=0.707 I _0$
It is clear from expression (1) that root mean square value of alternating current is $70.7 \%$ of its peak value.

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Question 32 Marks

When a capacitor is joined with a series LR circuit in series, then current flowing through the circuit increases. Explain, why?

Answer

Impedance of L-R circuit
$ Z_1=\sqrt{R^2+(\rho L)^2} \quad
\therefore \text { Current } I_1=\frac{E}{Z_1} $
On joining capacitor, impedance of LCR circuit
$Z _2=\sqrt{ R ^2+\left(\omega L -\frac{1}{\omega C }\right)^2} \therefore$ Current $I _2=\frac{ E }{ Z _2}$
$\therefore \quad \frac{ I _2}{ I _1}=\frac{ Z _1}{ Z _2}$
Since, $Z_1>Z_2$ Hence $\quad I _2> I _1$

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Question 42 Marks

A series LCR circuit with R = 20 Ω L = 1.5 H and $C=35 \mu F$ is connected to a variable frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Answer

Given : $R =20 \Omega, L=1.5 H$
$C =35 \mu F=35 \times 10^{-6} F$
$E _{ rms }=200 V$
Frequency, $f=$ variable
When the frequency of the supply is equal to the natural frequency of the circuit, then resonance occurs and in this condition$X_L=X_C$,
In this condition $Z=R=20$ and $\phi=0^{\circ}$
$\because \quad Z =\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2}$
$I_{ rms }=\frac{E_{ rms }}{Z}=\frac{200}{20}=10 A$
Power transferred in one complete cycle
$= E _{ rms } \times I _{ rms }$
$P =200 V \times 10 A$
$=2000 W=2 kW$

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Question 52 Marks

A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillation of the circuit?

Answer

Given :$C=30 uF=30 \times 10^{-6} F$
$L =27 mH =27 \times 10^{-3} H$
$\omega_r=?$
We know that $\quad \omega_r=\frac{1}{\sqrt{ LC }}$
or$\omega_r=\frac{1}{\sqrt{27 \times 10^{-3} \times 30 \times 10^{-6}}}$
$=\frac{1}{\sqrt{81 \times 10^{-8}}}=\frac{1}{9 \times 10^{-4}}$
$\omega_r=\frac{10^4}{9}=1.1 \times 10^3 rad s ^{-1}$

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Question 62 Marks

In questions 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.

Answer

For capacitance and induction coil, the phase difference $\phi$ between voltage and current is $\pm \frac{\pi}{2}$
Power absorbed in one cycle,
$P=E_{rms} I_{rms} \cos \phi$
$P=E_{r m s} I_{r m s} \cos \left( \pm \frac{\pi}{2}\right)=0$
Here,$\phi=\frac{\pi}{2}$
$\therefore \quad \cos \phi=\cos \frac{\pi}{2}=0$
$\therefore \quad P _{ av }=0$ in all the conditions.

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Question 72 Marks

A 44 m H inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Answer

Given : $\quad L =44 mH =44 \times 10^{-3} H$
$E _{ rms }=220 V$
$f=50 Hz$
$I _{ rms }=?$
We know that, $\quad X_L=L \omega$
$= L \times 2 \pi f$
$X _{ L }=44 \times 10^{-3} \times 2 \times \frac{22}{7} \times 50 \Omega$
$\therefore \quad I _{ rms }=\frac{ E _{ rms }}{ X _{ L }}$
On putting values, $I _{ rms }=\frac{220 \times 7}{44 \times 10^{-3} \times 2 \times 22 \times 50}$
$=\frac{220 \times 7 \times 10^3}{44 \times 2 \times 22 \times 50}=\frac{7 \times 10^3}{44 \times 10}$
$=\frac{700}{44}=\frac{175}{11}$
$=15.9 A$

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Question 82 Marks

(a) The peak voltage of an ac supply is 300 V. What is its rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?

Answer

Given: $\quad E _0=300 V$
$E _{ rms }=?$
$\begin{aligned} I _{ rms } & =10 A \\
I _0 & =?\end{aligned}$
(a) We know that,
$E _{ rms }=\frac{ E _0}{\sqrt{2}}=0.707 E _0$
$=0.707 \times 300$
$=212.100$
(b)$I_{rms}=\frac{I_0}{\sqrt{2}}$
$\therefore \quad I _0=\sqrt{2} I _{ rms }$
$=\sqrt{2} \times 10$
$=1.414 \times 10=14.14 A$

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