Physics

7.7 Capacitance of the capacitor, C = 30 $\mu F$ = 30 $*{10}^{-6}$ F

Inductance of the Inductor, L = 27 mH = 27 $*{10}^{-3}$ H

Angular frequency is given as

${\omega }_r$ = $\frac{1}{\sqrt[]{LC}}$ = $\frac{1}{\sqrt[]{27\mathrm{}*{10}^{-3}*30\mathrm{}*{10}^{-6}}}$ = 1.11 ${10}^3$ rad /s

7.6 Inductance, L = 2.0 H

Capacitance, C = 32 $\mu F$ = 32 $*{10}^{-6}$ F

Resistance, R = 10 Ω

Resonant frequency is given by the relation,

${\omega }_r$ = $\frac{1}{\sqrt[]{LC}}$ = $\frac{1}{\sqrt[]{2*32\mathrm{}*{10}^{-6}}}$ = 125 rad /s

Now Q value of the circuit is given as

Q = $\frac{1}{R}\sqrt[]{\frac{L}{C}}$ = $\frac{1}{10}\sqrt[]{\frac{2}{32\mathrm{}*{10}^{-6}}}$ = 25

7.5 In the inductive circuit:

rms value of current, I = 15.92 A

rms value of voltage, V = 220 V

The net power absorbed can be obtained by

P = VI cos $\varnothing$ , where $\varnothing$ is the phase difference between alternating voltage and current = 90 $°$

So P = VI cos $90°$ = 0

In the capacitive circuit:

rms value of current, I = 2.49 A

rms value of voltage, V = 110 V

The net power absorbed can be obtained by

P = VI cos $\varnothing$ , where $\varnothing$ is the phase difference between alternating voltage and current = 90 $°$

So P = VI cos $90°$ = 0

7.4 Capacitance of the capacitor, C = 60 $\mu F$ = 60 $*{10}^{-6}$ F

Supply voltage, V = 110 V

Supply frequency,  $\nu$ = 60 Hz

Angular frequency,  $\omega$ = 2 $\pi \nu$

Capacitive reactance,  $X_c$ = $\frac{1}{\omega C}$ = $\frac{1}{2\pi \nu C}$

rms value of current is given by I = $\frac{V}{X_c}$ = 110 $*2*\pi *60*$ 60 $*{10}^{-6}$ = 2.49 A

Hence, the rms value of current is 2.49 A.

7.3 Inductance of the Inductor, L = 44 mH = 44 $*{10}^{-3}$ H

Supply voltage, V = 220 V

Supply frequency,  $\nu$ = 50 Hz

Angular frequency,  $\omega$ = 2 $\pi \nu$

Inductive reactance,  $X_L$ = $\omega L$ = 2 $\pi \nu L$

rms value of current is given as

I = $\frac{V}{X_L}$ = $\frac{220}{2\pi *50*44\mathrm{}*{10}^{-3}}$ = 15.92 A

Hence, the rms value of the current in the circuit is 15.92 A.

7.2 Peak voltage of the AC supply,  $V_o$ = 300 V

RMS voltage is given by $V_{rms}$ = $\frac{\mathrm{}{V}_o}{\surd 2}$ = $\frac{\mathrm{}300}{\surd 2}$ = 212.13 V

The rms value of current, I = 10 A

Peak current $I_o$ = $\sqrt[]{2}I$ = $\sqrt[]{2}*10=14.1A$

7.1 Resistance of the resistor, R = 100 Ω

Supply voltage, V = 220 V

Supply frequency,  $\nu$ = 50 Hz

The rms value of current is given as, I = $\frac{V}{R}$ = $\frac{220}{100}$ = 2.2 A

The net power consumed over a full cycle is given as P = VI = 220 $*2.2$ = 484 W

6.17 Line charge per unit length = $\lambda$ = $\frac{Totalcharge}{Length}$ = $\frac{Q}{2\pi r}$

Where r = distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

Magnetic field, $\stackrel{?}{B}$ = $-B_0\stackrel{?}{k}$

At a distance r, the magnetic force is balanced by the centripetal force. i.e.

BQ $v$ = $\frac{M{v}^2}{r}$ , where v = linear velocity of the wheel. Then,

B $*2\lambda \pi r$ = $\frac{Mv}{r}$

v = $\frac{2B\lambda \pi r^2}{M}$

Angular velocity, $\omega =\frac{v}{R}$ = $\frac{2B\lambda \pi r^2}{MR}$

For r $\le a\le R,weget$ $\omega =-\frac{2B_0\lambda \pi a^2}{MR}\stackrel{?}{k}$

6.15 Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25 ${cm}^2$ = 25 $*{10}^{-4}$ $m^2$

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Current flowing time, t = ${10}^{-3}$ s

We know average back emf, e = $\frac{d\varnothing }{dt}$ …………………. (1)

Where $d\varnothing =$ Change in flux = NAB ……………… .(2)

B = Magnetic field strength = ${\mu }_0\frac{NI}{l}$ ………………. (3)

Where ${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$

From equation (1) and (2), we get

e = $\frac{NAB}{dt}$ = $\frac{NA}{dt}*{\mu }_0\frac{NI}{l}$ = $\frac{{\mu }_0{N}^{2}AI}{lt}$ = $\frac{4\pi *{10}^{-7}*{500}^2*25\mathrm{}*{10}^{-4}*2.5}{0.3*{10}^{-3}}$ = 6.544 V

Hence the average back emf induced in the solenoid is 6.5 V