Physics Ncert Solutions Class 12th

7.16 Capacitance of the capacitor, C = 100 $\mu F$ = 100 $*{10}^{-6}$ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Frequency of the supply voltage, $\nu$ = 12 kHz = 12 $*{10}^3$ Hz

Angular frequency, $\omega$ = 2 $\pi \nu$ = 2 $\pi *12\mathrm{}*{10}^3$ rad/s = 24 $\pi *{10}^3$ rad/s

For a RC circuit, the impedance Z, is given by Z = $\sqrt[]{R^2+\frac{1}{{\omega }^2{C}^2}}$

Z = $\sqrt[]{{40}^2+\frac{1}{{(24\pi *{10}^3)}^2{*(100*{10}^{-6})}^2}}$ = 40 Ω

Peak voltage, $V_0$ = $\surd 2*V$ = $\surd 2*110$ = 155.56 V

Peak current, $I_0$ = $\frac{V_0}{Z}$ = $\frac{155.56\mathrm{}}{40}$ = 3.8

9 A

In a capacitor circuit, the voltage lags behind the current by a phase angle of $\varnothing$ . This angle is given by the relation

$\tan ?\varnothing$ = $\frac{\frac{1}{\omega C}}{R}$ = $\frac{1}{\omega CR}$ = $\frac{1}{24\pi *{10}^3*100*{10}^{-6}*40}$ = 3.315 $*$ ${10}^{-3}$

$\varnothing =0.19°=\frac{0.19°*\pi }{180}$ rad

$\mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{a}\mathrm{g},t=\frac{\varnothing }{\omega }$ = $\frac{\frac{0.19°*\pi }{180}}{24\pi *{10}^3}$ = $\frac{0.19*\pi }{24\pi *{10}^3*180}$ = 4.4 $*{10}^{-8}$ s = 0.044 $\mu$ s

Hence, $\varnothing$ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor.

In DC circuit, after the steady state is achieved, $\omega$ = 0

Hence, capacitor C amounts to an open circuit.

7.15 Capacitance of the capacitor, C = 100 $\mu F$ = 100 $*{10}^{-6}$ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Frequency of the supply voltage, $\nu$ = 60 Hz

Angular frequency, $\omega$ = 2 $\pi \nu$ = 2 $\pi *60$ rad/s

For a RC circuit, the impedance Z, is given by Z = $\sqrt[]{R^2+\frac{1}{{\omega }^2{C}^2}}$

Z = $\sqrt[]{{40}^2+\frac{1}{{(2\pi *60)}^2{*(100*{10}^{-6})}^2}}$ = 47.996 Ω

Peak voltage, $V_0$ = $\surd 2*V$ = $\surd 2*110$ = 155.56 V

Peak current, $I_0$ = $\frac{V_0}{Z}$ = $\frac{155.56\mathrm{}}{47.996}$ = 3.24 A

In a capacitor circuit, the voltage lags behind the current by a phase angle of $\varnothing$ . This angle is given by the relation

$\tan ?\varnothing$ = $\frac{\frac{1}{\omega C}}{R}$ = $\frac{1}{\omega CR}$ = $\frac{1}{2\pi *60\mathrm{}*100*{10}^{-6}*40}$ = 0.663

$\varnothing =33.55°=\frac{33.55°*\pi }{180}$ rad

$\mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{a}\mathrm{g},t=\frac{\varnothing }{\omega }$ = $\frac{\frac{33.55°*\pi }{180}}{2\pi *60}$ = $\frac{33.55*\pi }{2\pi *60*180}$ = 1.55 $*{10}^{-3}$ s = 1.55 ms

Hence the time lag between maximum current and maximum voltage is 1.55 ms.

7.14 Inductance of the Inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of supply voltage, V = 240 V

Frequency of the supply, $\nu$ = 10 kHz = 10 $*{10}^3$ Hz

Peak voltage is given as $V_0$ = $\sqrt[]{2V}=\sqrt[]{2}*240$ = 339.41 V

Angular frequency of the supply, $\omega =2\pi \nu$ = 2 $\pi *{10}^{4}rad/s$

Maximum current in the supply is given as

$I_0$ = $\frac{V_0}{\sqrt[]{R^2+{\left(\omega L\right)}^2}}$ = $\frac{339.41}{\sqrt[]{{100}^2+{(2\pi *{10}^4*0.50)}^2}}$ = 10.80 $*{10}^{-3}$ A

Phase angle is also given by the relation,

$\tan ?\varnothing$ = $\frac{\omega L}{R}$ = $\frac{2\pi *{10}^4*0.5}{100}$ = 314.16

$\varnothing =89.82°$

= $\frac{89.82*\pi }{180}$ rad = 1.568 rad

$\omega t=1.568$

t = $\frac{1.568}{\omega }$ = $\frac{1.568}{2\pi *{10}^4}$ = 24.95 $*{10}^{-6}$ s = 24.95 $\mu$ s

It can be observed that $I_0$ is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit.

In a DC circuit, after a steady state is achieved, $\omega =0.$ Hence, inductor L behaves like a pure conducting object.

7.13 Inductance of the Inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of supply voltage, V = 240 V

Frequency of the supply, $\nu$ = 50 Hz

Peak voltage is given as $V_0$ = $\sqrt[]{2V}=\sqrt[]{2}*240$ = 339.41 V

Angular frequency of the supply, $\omega =2\pi \nu$ = 2 $\pi *50=314.16rad/s$

Maximum current in the supply is given as

$I_0$ = $\frac{V_0}{\sqrt[]{R^2+{\left(\omega L\right)}^2}}$ = $\frac{339.41}{\sqrt[]{{100}^2+{(314.16*0.50)}^2}}$ = 1.82 A

Equation for voltage is given as V = $V_0$ $\cos ?\omega t$ and equation for current is given as

I = $I_0\cos ?(\omega t-\varnothing )$ , where $\varnothing =$ phase difference between voltage and current.

At time t = 0, V = $V_0$ [Maximum voltage condition]

For $\omega t-\varnothing$ = 0, I = $I_0$ [ Maximum current condition], for that t = $\frac{\varnothing }{\omega }$

Hence the time lag between maximum voltage and maximum current is $\frac{\varnothing }{\omega }$

Phase angle is also given by the relation,

$\tan ?\varnothing$ = $\frac{\omega L}{R}$ = $\frac{314.16*0.5}{100}$ = 1.5708

$\varnothing =57.52°$

= $\frac{57.52*\pi }{180}$ rad = 1.003 rad

$\omega t=1.003$

t = $\frac{1.003}{\omega }$ = $\frac{1.003}{314.16}$ = 3.20 $*{10}^{-3}$ s = 3.2 ms

Hence the time lag between maximum voltage and maximum current is 3.2 ms.

7.12 Inductance of the Inductor, L = 20 mH = 20 $*$ ${10}^{-3}$ H

Capacitance of the capacitor, C = 50 $\mu F$ = 50 $*{10}^{-6}$ F

Initial charge of the capacitor, Q = 10 mC = 10 $*{10}^{-3}$ C

The total energy stored initially at the circuit is given as

E = $\frac{1}{2}$ $*\frac{Q^2}{C}$ = $\frac{1}{2}$ $*\frac{{(10\mathrm{}*{10}^{-3})}^2}{50\mathrm{}*{10}^{-6}}$ = 1 J

Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.

Natural frequency of the circuit is given by the relation

$\nu =\frac{1}{2\pi \sqrt[]{LC}}$ = $\frac{1}{2\pi \sqrt[]{20*\mathrm{}{10}^{-3}*50\mathrm{}*{10}^{-6}}}$ = 159.15 Hz

Natural angular frequency, $\omega =\frac{1}{\sqrt[]{LC}}$ = $\frac{1}{\sqrt[]{20*\mathrm{}{10}^{-3}*50\mathrm{}*{10}^{-6}}}$ = 1000 rad/s

(i) Total time period, T = $\frac{1}{\nu }$ = $\frac{1}{159.15}$ = 6.28 $*{10}^{-3}$ s = 6.28 ms

Total charge on the capacitor at time t is given by

Q' = Q $\cos ?\frac{2\pi }{T}$ t

For energy stored in electrical, we can write Q = Q'

Hence, it can be inferred that the energy stored in the capacitor is completely electrical at time, t = 0, $\frac{T}{2}$ , T , $\frac{3T}{2}$ , ……

(ii) Magnetic energy is the maximum when electrical energy Q' s equal to zero.

Hence it can be inferred that the energy stored in the capacitor is completely magnetic at time, t = $\frac{T}{4},\frac{3T}{4},\frac{5T}{4}$ …….

Let Q' be the charge on the capacitor when total energy is equally shared between the capacitor and the inductor a time t. When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor = $\frac{1}{2}$ (maximum energy)

= $\frac{1}{2}\frac{{Q\text{'}}^2}{C}$ = $\frac{1}{2}$ ( $\frac{1}{2}\frac{Q^2}{C}$ ) = $\frac{1}{4}\frac{Q^2}{C}$

${Q\text{'}}^2$ = $\frac{Q^2}{2}$ or Q' = $\frac{Q}{\surd 2}$

We have previously obtained Q' = Q $\cos ?\frac{2\pi }{T}$ t

$\frac{Q}{\surd 2}=$ Q $\cos ?\frac{2\pi }{T}$ t

$\cos ?\frac{2\pi }{T}$ t = $\frac{1}{\surd 2}$ = cos(2 $n+1)\frac{\pi }{4}$ , where n = 0,1,2,3…….

t = (2n+1) $\frac{T}{8}$

Hence, total energy is equally shared between the inductor and the capacitor at time,

t = $\frac{T}{8}$ , $\frac{3T}{8}$ , $\frac{5T}{8}$ ,………

Resistor damps out the LC oscillations eventually. The whole of the initial energy (= 1.0 J) is eventually dissipated as heat.

7.11 Inductance of the Inductor, L = 5.0 H

Resistance of the resistor, R = 40 Ω

Capacitance of the capacitor, C = 80 $\mu F=$ 80 $*{10}^{-6}$ F

Potential of the voltage source, V = 230 V

Resonance angular frequency is given as

${\omega }_r$ = $\frac{1}{\sqrt[]{LC}}$ = $\frac{1}{\sqrt[]{5*80\mathrm{}*{10}^{-6}}}$ = 50 rad/s

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{i}\mathrm{r}\mathrm{c}\mathrm{u}\mathrm{i}\mathrm{t}\mathrm{}\mathrm{w}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{n}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}\mathrm{a}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}50\mathrm{}\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}$

The impedance of the circuit is given as

Z = $\sqrt[]{R^2+{\left(X_{L-}{X}_C\right)}^2}$ where $X_L$ = Inductive reactance and $X_C$ = Capacitive reactance

At resonance, $X_L$ = $X_C$ so Z = R = 40Ω

Amplitude of the current at the resonating frequency is given as

$I_o$ = $\frac{V_0}{Z}$ where $V_0$ = Peak voltage = $\surd 2$ V

$I_o$ = $\frac{\surd 2\mathrm{V}}{Z}$ = $\frac{\surd 2*230}{40}$ = 8.13 A

Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.

rms potential drop across the inductor is given by

${{(V}_l)}_{rms}$ = $I_{rms}$ $*{\omega }_{r}L$

$I_{rms}=$ $\frac{I_o}{\surd 2}$ = $\frac{8.13}{\surd 2}$ = 5.75 A

$Hence,$ ${{(V}_l)}_{rms}$ = $5.75$ $*50*5$ = 1437.5 V