5 Marks Questions

Question 15 Marks

$a.$ Derive the expression for the current flowing in an ideal capacitor and its reactance when connected to an ac source of voltage $V = V _{ o } \sin \omega t$.
$b.$ Draw its phasor diagram.
$c.$ If resistance is added in series to capacitor what changes will occur in the current flowing in the circuit and phase angle between voltage and current.

Answer

$a.$ We have $V = V _{ o } \sin \omega t$.
Also, $v =\frac{q}{c} ; q =$ charge on capacitor
$v _0 \sin \omega t=\frac{q}{c}$
or, $q = cv _0 \sin \omega t$
$\therefore I =\frac{d q}{d t}=\frac{d}{d t}\left( CV _0 \sin \omega t\right)= cv _0 \sin \omega t \cdot \omega$
$\therefore I =\frac{ v _0}{\frac{1}{\omega t}} \sin \left(\omega t+\frac{\pi}{2}\right)$
Max. current, $I _{ o }=\frac{ v _o}{1} \times 1$ when $\sin \left(\omega t+\frac{\pi}{2}\right)=1$
Comparing with ohm's law: $I =\frac{V}{R}$ to equation $I _{ o }=\frac{ v _o}{\frac{1}{ uc }}$
We have, capacitive reactance, $x _{ C }=\frac{1}{\omega c}$
$b.$ Phasor diagram:

$c.$ A resistor is now connected with the capacitor in series:

Peak voltage drop across $R$ is $i_0 R$
Peak voltage drop across $C$ is $i_0 X_C$.
Voltage a cross $R$ is in phase with the current.
Voltage across $C$ lags the current by $90^{\circ}$.
So, the voltage drops across $R$ and across $C$ are not in phase.
They are out of phase by $90^{\circ}$.
$\text { So, } \varepsilon_0=\sqrt{\left(i_0 R\right)^2+\left(i_0 X_C\right)^2}$
$\therefore i _0=\frac{V_0}{\sqrt{R^2+X_C^2}}$
The phase angle is

Phase Angle $=\phi=\tan ^{-1} \frac{X_C}{R}$

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Question 25 Marks

You have learned in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.

Answer

We are given a plane mirror $XY$ and let, $O$ be a point object at a distance $OP,$ in front of the plane mirror.
$A$ part $\text{RPQ}$ of the wavefront touches the plane mirror at $P$ and from this point, spherical wavefronts start emanating.
Whereas disturbance from $R$ and $Q$ continues moving forward, along with the normal rays $OR$ and $OQ,$ that reflects back $v$.
When, disturbances from $R, P,$ and $Q$ reach the mirror at $A, B\ '$ and $C$ respectively, reflected spherical wavefront is formed.

The reflected wavefront $AB'C$ appears to start from $I$.
Hence $,I$ become a virtual image for $0$ as a real point object.
Draw $AN$ normal to $XY,$ hence parallel to $OP$.
Now $,OA$ is the incident ray $($being normal to incident wavefront $\text{ABC})$ and $AD$ is the reflected ray $($being normal to reflected wavefront $AB'C).$
Thus, $\angle OAN =\angle DAN =\theta \ [ i = r ]$
But, $\angle O A N=$ alternate $\angle AOP$
and $\angle DAN =$ corresponding $\angle AlP$
$\therefore \angle AOP =\angle AlP$
$\angle AIP =\angle AOP \ ($ each $ \theta)$
$\angle A P=\angle A P O=90^{\circ} \ ($each $90^{\circ})$
$AP$ is common to both
$\Delta_3$ become congruent
Hence $, PI = PQ$
i.e., a normal distance of the image from the mirror $=$ normal distance of the object from the mirror.
Thus, a virtual image is formed as much behind the mirror as the object in front of it.

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Question 35 Marks

Draw a ray diagram to show the formation of real image of the same size as that of the object placed in front of a converging lens. Using this ray diagram establish the relation between u, v and f for this lens.

Answer

Thin Lens Formula : Suppose an object AB of finite size is placed normally on the principal axis of a thin convex lens (fig.). A ray AP starting from A parallel to the principal axis, after refraction through the lens, passes through the second focus F. Another ray AC directed towards the optical centre C of the lens, goes straight undeviated. Both the rays meet at A' Thus A' is the real image of A. The perpendicular A' B' dropped from A' on the principal axis is the whole image of AB.

Let distance of object AB from lens = u
Distance of image A' B' from lens = v
Focal length of lens = f. We can see that triangles ABC and A' B' C' are similar
$\frac{A B}{A^{\prime} B^{\prime}}=\frac{C B}{C B^{\prime}} \ldots$ (i)
Similarly triangles PCF and A' B' F are similar
$\frac{P C}{A^{\prime} B^{\prime}}=\frac{C F}{F B^{\prime}}$
But PC = A B
$\frac{A B}{A^{\prime} B^{\prime}}=\frac{C F}{F B^{\prime}} \ldots$ (ii)
From (i) and (ii), we get $\frac{C B^{\prime}}{C B^{\prime}}=\frac{C F^{\prime}}{F B^{\prime}} \ldots$ (iii)
From sign convention, $CB =- u , CB ^{\prime}= v , CF = f$ and $FB ^{\prime}= CB ^{\prime}- CF = v - f$
Substituting this value in (iii), we get, $-\frac{u}{v}=\frac{f}{v-f}$
or -u(v - f) = vf or -uv + uf = vf
Dividing throughout by uvf, we get $\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \ldots$ (iv)

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Question 45 Marks

$i$. An ac source generating a voltage $V = V _0 \sin \omega t$ is connected to a capacitor of capacitance $C$ . Find the expression of the current $I$ flowing through it. Plot a graph of $V$ and $I$ versus $\omega t$ to show that the current is $\frac{\pi}{2}$ ahead of the voltage.
$ii$. A resistor of $200 \Omega$ and a capacitor of $15 \mu F$ are connected in series to a $220 V, 50 Hz$ ac source. Calculate the current in the circuit and the rms voltage across the resistor and the capacitor. Why the algebraic sum of these voltages is more than the source voltage?

Answer

$i. V = V _0 \sin \omega t, V=\frac{Q}{C}$
$A.C$. source containing capacitor:
Let a source of alternating $\text {e.m.f.}$
$V = V _{ m } \sin \omega t$ be connected to a capacitor of capacitance $C$ only.
$l=\frac{d Q}{d t}$
$I_0=\frac{V_0}{\left(\frac{1}{\omega C}\right)}$
$I = I _0 \sin \left(\omega t+\frac{\pi}{2}\right)$

$ii. X_C=\frac{1}{2 \pi / C}=212.3 \Omega$
$ Z =\sqrt{R^2+X_C^2}=291.5 \Omega$
$I_{r m s}=\frac{V_{r m s}}{Z}=\frac{220}{2915}=0.755 A$
$V_R(ms)=151 V$
$V_C( mss )=160.3 V$
Two voltages are out of phase, hence they are added vectorially.

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Question 55 Marks

Define the terms (i) capacitance of a capacitor (ii) dielectric strength of a dielectric. When a dielectric is inserted between the plates of a charged parallel plate capacitor, fully occupying the intervening region, how does the polarization of the dielectric medium affect the net electric field? For linear dielectrics, show that the introduction of a dielectric increases its capacitance by a factor $\kappa$, characteristic of the dielectric.

Answer

i. Capacitance of a capacitor is defined as the ratio of the electric charge on the capacitor to the electric potential of capacitor due to it's charge.
ii. Dielectric strength of a dielectric is defined as the maximum value of electric field that can be applied to the dielectric without it's electric breakdown.
When a dielectric slab is introduced in between the plates of capacitor, the electric field gets reduced.
Consider a parallel plate capacitor with vacuum in between it's plates. The capacitor is charged up with battery such that electric field is set up between it's plates.
Then, $E^{\circ}=\sigma \varepsilon^{\circ}$
where, $\sigma$ is the surface charge density of the plates.
Now, as soon as the dielectric is introduced in between the plates each molecule of the dielectric get's polarised. Charges are induced on the surface of the dielectric and, these induced charges set up an electric field $E _{ p }$ inside the dielectric.
Therefore, the resultant electric field gets reduced and is given as
$E = E ^{ O }- E _{ p }$
When there is vaccum in between the plates, capacitance is given by $C =\varepsilon^{\circ} Ad$.
When, dielctric is inserted in between the plates, Capacitance increases by a factor of K.
where, K is the dielectric constant.
Capacitance becomes $C = K \varepsilon^{\circ}$ Ad.

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Question 65 Marks

A capacitor of capacitance $C_1$ is charged to a potential $V_1$ while another capacitor of capacitance $C_2$ is charged to a potential difference $V_2$. The capacitors are now disconnected from their respective charging batteries and connected in parallel to each other.
$i$. Find the total energy stored in the two capacitors before they are connected.
$ii$. Find the total energy stored in the parallel combination of two capacitors.
$iii.$ Explain the reason for the difference of energy in parallel combination in comparison to the total energy before they are connected

Answer

$i$. Total energy stored in the two capacitors before they are connected,
$u_i=\frac{1}{2} C_1 V_1^2+\frac{1}{2} C_2 V_2^2$
$ii$. After the two capacitors are connected in parallel, the common potential is
$V=\frac{\text { Total charge }}{\text { Total capacitance }}=\frac{q_1+q_2}{C_1+C_2}$
$=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}$
Total energy stored in the parallel combination,
$U_f=\frac{1}{2}\left(C_1+C_2\right) V^2$
$=\frac{1}{2}\left(C_1+C_2\right)\left(\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\right)^2$
$=\frac{1}{2} \frac{\left(C_1 V_1+C_2 V_2\right)^2}{C_1+C_2}$
$iii.$ Clearly, $U _{ f }< U _{ i }$.
Thus the total energy of the parallel combination is less than the sum of the energies stored in the two capacitors before they are connected.
During sharing of charges, some energy is lost as heat due to the flow of charges in connecting wires.

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