Class 12 Engineering Physics - Alternating Current

An $$AC$$ source generating a voltage $$e = e_{0}\sin \omega t$$ is connected to a capacitor of capacitance $$C$$. Find the expression for the current $$i$$ flowing through it. Plot a graph of $$e$$ and $$i$$ versus $$\omega t$$.

$$e=e_{0}\sin \omega t$$ $$e=\dfrac{Q}{C}$$

$$i=\dfrac{dQ}{dt}$$

$$i_{0}=\dfrac{e_{0}}{\left(\dfrac{1}{\omega C}\right)}$$

$$i=i_{0}\sin \left(\dfrac{\omega t+\pi}{2}\right)$$

1766128_1842200_ans_7c0bfb2faf7a464bb896c24cbea48378.png

In the double-slit experiment of Fig. $$ 35-10 $$, the electric fields of the waves arriving at point $$ P $$ are given by
$$E_{1}=(2.00 \mu \mathrm{V} / \mathrm{m}) \sin \left[\left(1.26 \times 10^{15}\right) t\right]$$
$$E_{2}=(2.00 \mu \mathrm{V} / \mathrm{m}) \sin \left[\left(1.26 \times 10^{15}\right) t+39.6 \ \mathrm{rad}\right]$$
where time $$ t $$ is in seconds.
In a phasor diagram of the electric fields, what is the angle between the phasors?

The angle between the phasors is $$ \phi=39.6 \ \mathrm{rad}=2270^{\circ} $$ (which would look like about $$ 110^{\circ} $$ when drawn in the usual way)

What transformation of energy takes place when current is drawn from a cell ?

When current is drawn from a cell, chemical energy is converted into electrical energy.

Write phase difference between current and voltage for an ac circuit, when (I) $$f={f}_{r}$$ (ii) $$f< {f}_{r}$$ (iii) $$f> {f}_{r}$$ here $${f}_{r}$$ is the resonant frequency.

(i) $$f={f}_{r}$$ when $${X}_{L}={X}_{C}$$
The circuit is purely resistive ie., it is non-inductive. Therefore, current and voltage in the circuit are in the same phase.
(ii) $$f< {f}_{r}$$ as $${ X }_{ L }=\omega L=2\pi fL;{ X }_{ C }=\cfrac { 1 }{ \omega C } =\cfrac { 1 }{ 2\pi fC } \quad $$
When $$f$$ is small, $${X}_{L}$$ is small, $${X}_{C}$$ is large. The circuit is capacitance dominated. Therefore, current in the circuit leads the voltage by phase angle $$\phi$$
(iii) $$f> {f}_{r}$$
So, $${X}_{L}$$ becomes large and $${X}_{C}$$ becomes small. The circuit is inductance dominated. The current lags behind the voltage by phase angle $$\phi$$

Add the quantities $$ y_{1}=10 \sin \omega t, y_{2}=15 \sin \left(\omega t+30^{\circ}\right) $$ and $$ y_{3}=5.0 \sin \left(\omega t-45^{\circ}\right) $$ using the phasor method.

In adding these with the phasor method (as opposed to, say, trig identities), we may set $$ t=0 $$ and add them as vectors
$$y_{h}=10 \cos 0^{\circ}+15 \cos 30^{\circ}+5.0 \cos \left(-45^{\circ}\right)=26.5 $$
$$y_{v}=10 \sin 0^{\circ}+15 \sin 30^{\circ}+5.0 \sin \left(-45^{\circ}\right)=4.0$$
so that $$y_{R}=\sqrt{y_{h}^{2}+y_{v}^{2}}=26.8 \approx 27$$
$$\beta=\tan ^{-1}\left(\dfrac{y_{v}}{y_{h}}\right)=8.5^{\circ} $$
$$\text { Thus, } y=y_{1}+y_{2}+y_{3}=y_{R} \sin (\omega t+\beta)=27 \sin \left(\omega t+8.5^{\circ}\right)$$

In the double-slit experiment of Fig. $$ 35-10 $$, the electric fields of the waves arriving at point $$ P $$ are given by
$$E_{1}=(2.00 \mu \mathrm{V} / \mathrm{m}) \sin \left[\left(1.26 \times 10^{15}\right) t\right]$$
$$E_{2}=(2.00 \mu \mathrm{V} / \mathrm{m}) \sin \left[\left(1.26 \times 10^{15}\right) t+39.6 \mathrm{rad}\right]$$
where time $$ t $$ is in seconds.
In a phasor diagram of the electric fields, at what rate would the phasors rotate around the origin?

The rate is given by $$ \omega=1.26 \times 10^{15} \mathrm{rad} / \mathrm{s} $$

Find the sum $$ y $$ of the following quantities:
$$y_{1}=10 \sin \omega t \quad \text { and } \quad y_{2}=8.0 \sin \left(\omega t+30^{\circ}\right)$$

In adding these with the phasor method (as opposed to, say, trig identities), we may set $$ t=0 $$ and add them as vectors:
$$y_{h}=10 \cos 0^{\circ}+8.0 \cos 30^{\circ}=16.9 $$
$$y_{v}=10 \sin 0^{\circ}+8.0 \sin 30^{\circ}=4.0$$
so that $$y_{R}=\sqrt{y_{h}^{2}+y_{v}^{2}}=17.4$$
$$\beta=\tan ^{-1}\left(\dfrac{y_{v}}{y_{b}}\right)=13.3^{\circ}$$
Thus,$$y=y_{1}+y_{2}=y_{R} \sin (\omega t+\beta)=17.4 \sin \left(\omega t+13.3^{\circ}\right)$$
Quoting the answer to two significant figures, we have $$ y \approx 17 \sin \left(\omega t+13^{\circ}\right) $$

Compare resistance and reactance.

Key Differences Between Resistance and Reactance.

1. The resistance is the obstacle in the flow of current in an electrical circuit due to resistor. While reactance is the opposition to the charging current due to either inductor or capacitor.

2. Resistance is the property associated with both ac and dc circuit. However, reactance to the property is only associated with ac circuits.

3. Pure resistors generate resistance. As against ideal inductors or capacitors give rise to reactance in the circuit.

4. Resistance is associated with the real part of the impedance. While reactance contributes to the imaginary part of the impedance value.

5. The difference in phase between voltage and current in a purely resistive circuit is $$0^{o}$$. While the phase difference between voltage and current in an ideal capacitive or inductive circuit is $$90^{o}$$. In the case of inductive load, the current lags behind the voltage by $$90^{o}$$ and for purely capacitive load voltage lag the current by $$90^{o}$$.

6. The resistance offered by the circuit depends on the dimension, resistivity, and temperature conditions of the conductor. However, the reactance relies on the frequency component of the alternating current in the circuit. It shows proportionality with frequency in case of inductive load, whereas this relation is inverse in case of capacitive load.

7. In the resistive circuit, the overall supplied power to the circuit gets dissipated in the form of heat. Whereas in a capacitive or inductive circuit, the device does not fully consume the total supplied power.

An $$AC$$ circuit consists of only an inductor of inductance $$2 H$$. If the current is represented by a sine wave of amplitude $$0.25 A$$ and frequency $$60 Hz$$, calculate the effective potential difference across the inductor $$(\pi = 3.142)$$

Here, $$L=2H$$,

$$I_{0}=0.25A$$,

$$v=60Hz$$

We know that:

$$V_{eff}=E_{eff}.X_{L}=\dfrac{I_{0}}{\sqrt{2}}(2\pi vL)$$

$$=\dfrac{0.25}{1.414}\times 2\times 3.142\times 60\times 2$$

$$= 133.32V$$

Draw the approximate voltage vector diagrams in the electric circuits shown in Fig.$$4.33$$ $$a,b.$$ The external voltage $$V$$ is assumed to e alternating harmonically with frequency $$\omega$$.

$$\tan \phi =\dfrac{\omega L-\dfrac{1}{\omega C}}{R}=-ve$$
as
$$\omega^2<\dfrac{1}{LC}$$
1787536_1853312_ans_ae83c80ef44540cb9ed2319b26fc5ab0.png

A capacitor with capacitance $$C$$ and coil with active resistance $$R$$ and inductance $$L$$ are connected in parallel to a source of sinusoidal voltage voltage of frequency $$\omega$$. Find the phase difference between the current fed to the circuit and the source voltage.

We use the method of complex voltage
$$V=V_0e^{i\omega t}$$
Then $$I_C=\dfrac{V_0e^{i\omega t}}{\dfrac{1}{i\omega C}}=i\omega CV_0e^{i\omega t}$$
$$I_{L,R}=\dfrac{V_0e^{i\omega t}}{R+i\omega L}$$
$$I=I_C+I_{L,R}=V_0\dfrac{R-i\omega L+i\omega C(R^2+\omega ^2L^2)}{R^2+\omega ^2L^2}e^{i\omega t}$$
Then taking the real part
$$I=\dfrac{V_0\sqrt{R^2+|\omega C(R^2+\omega ^2L^2)-\omega L|^2}}{R^2+\omega ^2L^2}\cos (\omega t-\phi)$$
Where
$$\tan \phi =\dfrac{\omega L-\omega C(R^2+\omega ^2L^2)}{R}$$

The power factor of an a.c. circuit is $$0.5$$ what is phase difference between voltage and current.

$$\textbf{Hint: Power factor=}$$$$cos\phi$$

Step-1: Define power factor

$${\text{ If }}\phi {\text{ be the phase difference between ac voltage (V) and current (I), }}$$

$${\text{Then power factor is defined by }}\cos \phi$$

Step-2: Calculate phase difference

$${\text{ Given ,}}$$

$${\text{ power factor cos}}\phi = 0.5$$

Therefore, Phase difference $$\phi$$ = $${\cos ^{ - 1}}(0.5) = 60^\circ$$

If a direct current of $$ \alpha $$ unit is superimposed with an alternating current $$ I = \beta sin \omega t $$ , then effective value of resulting current is

1356556_1587880_ans_8c9ff73d12b3409bb6cf8135c83524bc.jpg

Match the following column

i $$ L = \dfrac{\mu N^2 A }{l} $$
So, L depends upon shape , size ( 1 , A ) and medium $$ (\mu) $$ inserted

ii. $$ C = \dfrac{\varepsilon A}{d} $$
So, C depends upon shape , size (A , d) and medium(e) inserted

iii. $$ Z = \sqrt{R^2 + X^2_{L}} $$
When $$ x_l = \omega L , \omega $$ depends upon the external voltage source . So, Z depends upon all the factors in List II .

iv. $$ X_C = \dfrac{1}{\omega C} $$ (independent of resistivity )

An electrical device draws $$2kW$$ power from $$AC$$ mains $$[Voltage\ 223\ V (rms) = \sqrt {50,000}V]$$. The current differs (lags) in phase by $$\phi \left (\tan \phi = \dfrac {-3}{4}\right )$$ as compared to voltage. Find (i) $$R$$, (ii) $$X_{C} - X_{L}$$, and (iii) $$I_{M}$$. Another device has twice the values for $$R, X_{C}$$ and $$X_{L}$$. How are the answers affected?

$$P = 2000\ W$$ Current lags the voltage so
$$V^{2} = 50,000\ V\ X_{C} > X_{L}$$
$$\tan \phi = \dfrac {-3}{4}$$
$$I_{m} = I_{0}$$
$$P = \dfrac {V^{2}}{Z}$$
$$2000 = \dfrac {50,000}{Z}$$
$$Z = \dfrac {50,000}{2000} = 25\Omega$$
$$Z = \sqrt {R^{2} + (X_{C} - X_{L})^{2}}$$
$$R^{2} + (X_{C} - X_{L})^{2} = 25^{2} .... (I)$$
$$R^{2} + (X_{C} - X_{L})^{2} = 625$$
$$\tan \phi = \dfrac {-3}{4}$$
$$\dfrac {X_{C} - X_{L}}{R} = \dfrac {-3}{4}$$
$$X_{C} - X_{L} = \dfrac {-3}{4}R$$
$$(X_{C} - X_{L})^{2} = \dfrac {9}{16}R^{2} .... (II)$$
Put the value of $$(X_{C} - X_{L})^{2}$$ in (I)
$$R^{2} + \dfrac {9}{16} R^{2} = 625$$
$$\dfrac {25}{16}R^{2} = 625$$
$$R^{2} = 16\times 25$$
$$R = 4\times 5 = 20\Omega$$
$$\dfrac {X_{C} - X_{L}}{R} = \dfrac {-3}{4}$$
$$X_{C} - X_{L} = \dfrac {-3}{4} \times 20$$
$$X_{C} - X_{L} = -15\Omega$$
$$I_{rms} = \dfrac {V}{Z} = \dfrac {223}{25} \cong 9A$$
$$I_{0} = \sqrt {2} I_{rms} = \sqrt {2}\times 9$$
$$I_{0} = 12.6\ A$$
As (i) $$R, X_{C}, X_{L}$$ all are double $$\tan \tan \phi = \dfrac {X_{L} - X_{C}}{R}$$ will remain same. (ii) $$Z$$ will become double then $$I = \dfrac {V}{Z}$$ become half as value of $$V$$ does not change. (iii) As $$I$$ become half $$P = VI$$ will become again half as voltage remains same.

Suppose that a parallel-plate capacitor has circular plates
with radius $$R = 30$$ mm and plate separation of $$5.0$$ mm.
Suppose also that a sinusoidal potential difference with a maximum value of $$150$$ V and a frequency of $$60$$ Hz is applied across
the plates; that is,

$$V = (150 V) sin[2\pi_(60 Hz)t]$$

(a) Find $$B_{max}(R)$$, the maximum value of the induced magnetic field
that occurs at $$r = R$$. (b) Plot $$B_{max}(r)$$ for $$0 < r < 10\, cm$$.

(a) Noting that the magnitude of the electric field (assumed uniform) is given by

$$E =V/d$$ (where $$d = 5.0$$ mm), we use the result of part (a) in Sample Problem – “Magnetic
field induced by changing electric field:”

$$B=\dfrac{\mu_0\epsilon_0r}{2}\dfrac{dE}{dt}=\dfrac{\mu_0\epsilon_0r}{2d}\dfrac{dV}{dt}$$ ( $$r \leq R$$).

We also use the fact that the time derivative of $$\sin (ωt)$$ (where $$ω = 2πf = 2π(60) ≈ 377/s$$ in this problem) is $$ω\,cos(ωt)$$. Thus, we find the magnetic field as a function of r (for $$r ≤R$$; note that this neglects “fringing” and related effects at the edges):

$$B =\dfrac{\mu_0\epsilon_0r}{2d}V_{max}ω\,cos(ωt)\,\Rightarrow\,\,B_{max}=\dfrac{\mu_0\epsilon_0rV_{max}\omega}{2d}$$

where $$V_{max} = 150$$ V. This grows with r until reaching its highest value at $$r = R = 30$$ mm:

$$B_{max}\Bigg|_{r=R}=\dfrac{\mu_0\epsilon_0rV_{max}\omega}{2d}=\dfrac{(4\pi \times 10^{-7}\, H/m)(8.85 \times 10^{-12}\,F/m)(30 \times 10^{-3} \,m)(150 \,V)(377 \,s)}{2(5.0 \times 10^{-3}\, m)}$$

$$=1.9\times10^{-12}\,T$$

(b) For $$r \leq 0.03$$ m, we use the expression $$B_{max}=\mu_0\epsilon_0rV_{max}\omega/2d$$ found in part (a) (note
the $$B ∝ r$$ dependence), and for $$r \geq 0.03$$ m we perform a similar calculation starting with
the result of part (b) in Sample Problem — “Magnetic field induced by changing electric
field:”

$$B_{max}=\Bigg(\dfrac{\mu_0\epsilon_0R^2}{2r}\dfrac{dE}{dt}\Bigg)_{max}=\Bigg(\dfrac{\mu_0\epsilon_0R^2}{2rd}\dfrac{dV}{dt}\Bigg)_{max}=\Bigg(\dfrac{\mu_0\epsilon_0R^2}{2rd}V_{max}\omega\cos(\omega t)\Bigg)_{max}$$

$$=\Bigg(\dfrac{\mu_0\epsilon_0R^2V_{max}\omega}{2rd}\Bigg)_{max}$$ (for $$r \geq R$$)

(note the $$B ∝ r^{–1}$$ dependence — ). The plot (with SI units
understood) is shown below.

1650621_1777375_ans_80d4fe491d8e4ebf9b3bb41e8e695d25.jpg

The inductive current in the ring varies with time as $$I=I_m \sin\ (\omega t-\phi)$$, where $$I_m=\omega \phi _0\sqrt{R^2+\omega ^2L^2}$$ with $$tan\ \phi =\omega L/R$$.

we have $$\varepsilon =-\dfrac{d\Phi}{dt}=\omega \Phi_0\sin \omega t=LI+RI$$
Put
$$I=I_m \sin\ (\omega t-\phi)$$..Then
$$\omega \Phi_0\sin \omega t=\omega \Phi_0{\sin (\omega t-\phi)\cos \phi +\cos (\omega t-\phi)\sin \phi}$$
$$=LI_m \omega \cos (\omega t-\phi)+RI_m\sin (\omega t-\phi)$$
So
$$RI_m=\omega \Phi_0\cos \phi$$ and $$LI_m=\Phi_0\sin \phi$$
or
$$I_m=\dfrac{\omega \Phi_0}{\sqrt{R^2+\omega ^2L^2}}$$ and $$\tan \phi=\dfrac{\omega L}{R}$$

Additional Problems(81)
An AC source with $$\Delta = V_{rms} = 120 \,V$$ is connected between points a and $$d$$ in Figure. At what frequency will it deliver a power of $$250 \,W$$? Explain your answer.

1915239_1880068_ans_0e8ecc75bafc45dfa27a83b913c22c48.jpeg

Alternating Current - Class 12 Engineering Physics - Extra Questions

An $$AC$$ source generating a voltage $$e = e_{0}\sin \omega t$$ is connected to a capacitor of capacitance $$C$$. Find the expression for the current $$i$$ flowing through it. Plot a graph of $$e$$ and $$i$$ versus $$\omega t$$.

What transformation of energy takes place when current is drawn from a cell ?

Write phase difference between current and voltage for an ac circuit, when (I) $$f={f}_{r}$$ (ii) $$f< {f}_{r}$$ (iii) $$f> {f}_{r}$$ here $${f}_{r}$$ is the resonant frequency.

Add the quantities $$ y_{1}=10 \sin \omega t, y_{2}=15 \sin \left(\omega t+30^{\circ}\right) $$ and $$ y_{3}=5.0 \sin \left(\omega t-45^{\circ}\right) $$ using the phasor method.

Find the sum $$ y $$ of the following quantities:
$$y_{1}=10 \sin \omega t \quad \text { and } \quad y_{2}=8.0 \sin \left(\omega t+30^{\circ}\right)$$

Compare resistance and reactance.

An $$AC$$ circuit consists of only an inductor of inductance $$2 H$$. If the current is represented by a sine wave of amplitude $$0.25 A$$ and frequency $$60 Hz$$, calculate the effective potential difference across the inductor $$(\pi = 3.142)$$

Draw the approximate voltage vector diagrams in the electric circuits shown in Fig.$$4.33$$ $$a,b.$$ The external voltage $$V$$ is assumed to e alternating harmonically with frequency $$\omega$$.

A capacitor with capacitance $$C$$ and coil with active resistance $$R$$ and inductance $$L$$ are connected in parallel to a source of sinusoidal voltage voltage of frequency $$\omega$$. Find the phase difference between the current fed to the circuit and the source voltage.

The power factor of an a.c. circuit is $$0.5$$ what is phase difference between voltage and current.

If a direct current of $$ \alpha $$ unit is superimposed with an alternating current $$ I = \beta sin \omega t $$ , then effective value of resulting current is

Match the following column

The inductive current in the ring varies with time as $$I=I_m \sin\ (\omega t-\phi)$$, where $$I_m=\omega \phi _0\sqrt{R^2+\omega ^2L^2}$$ with $$tan\ \phi =\omega L/R$$.

Additional Problems(81)
An AC source with $$\Delta = V_{rms} = 120 \,V$$ is connected between points a and $$d$$ in Figure. At what frequency will it deliver a power of $$250 \,W$$? Explain your answer.

Class 12 Engineering Physics Extra Questions

Alternating Current - Class 12 Engineering Physics - Extra Questions

An $$AC$$ source generating a voltage $$e = e_{0}\sin \omega t$$ is connected to a capacitor of capacitance $$C$$. Find the expression for the current $$i$$ flowing through it. Plot a graph of $$e$$ and $$i$$ versus $$\omega t$$.

What transformation of energy takes place when current is drawn from a cell ?

Write phase difference between current and voltage for an ac circuit, when (I) $$f={f}_{r}$$ (ii) $$f< {f}_{r}$$ (iii) $$f> {f}_{r}$$ here $${f}_{r}$$ is the resonant frequency.

Add the quantities $$ y_{1}=10 \sin \omega t, y_{2}=15 \sin \left(\omega t+30^{\circ}\right) $$ and $$ y_{3}=5.0 \sin \left(\omega t-45^{\circ}\right) $$ using the phasor method.

Find the sum $$ y $$ of the following quantities:$$y_{1}=10 \sin \omega t \quad \text { and } \quad y_{2}=8.0 \sin \left(\omega t+30^{\circ}\right)$$

Compare resistance and reactance.

An $$AC$$ circuit consists of only an inductor of inductance $$2 H$$. If the current is represented by a sine wave of amplitude $$0.25 A$$ and frequency $$60 Hz$$, calculate the effective potential difference across the inductor $$(\pi = 3.142)$$

Draw the approximate voltage vector diagrams in the electric circuits shown in Fig.$$4.33$$ $$a,b.$$ The external voltage $$V$$ is assumed to e alternating harmonically with frequency $$\omega$$.

A capacitor with capacitance $$C$$ and coil with active resistance $$R$$ and inductance $$L$$ are connected in parallel to a source of sinusoidal voltage voltage of frequency $$\omega$$. Find the phase difference between the current fed to the circuit and the source voltage.

The power factor of an a.c. circuit is $$0.5$$ what is phase difference between voltage and current.

If a direct current of $$ \alpha $$ unit is superimposed with an alternating current $$ I = \beta sin \omega t $$ , then effective value of resulting current is

Match the following column

The inductive current in the ring varies with time as $$I=I_m \sin\ (\omega t-\phi)$$, where $$I_m=\omega \phi _0\sqrt{R^2+\omega ^2L^2}$$ with $$tan\ \phi =\omega L/R$$.

Additional Problems(81)An AC source with $$\Delta = V_{rms} = 120 \,V$$ is connected between points a and $$d$$ in Figure. At what frequency will it deliver a power of $$250 \,W$$? Explain your answer.

Class 12 Engineering Physics Extra Questions

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Find the sum $$ y $$ of the following quantities:
$$y_{1}=10 \sin \omega t \quad \text { and } \quad y_{2}=8.0 \sin \left(\omega t+30^{\circ}\right)$$

Additional Problems(81)
An AC source with $$\Delta = V_{rms} = 120 \,V$$ is connected between points a and $$d$$ in Figure. At what frequency will it deliver a power of $$250 \,W$$? Explain your answer.