Alternating Current

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

Potential drop across capacitor,

${{(V}_C)}_{rms}$ = $I_{rms}$ $*\frac{1}{{\omega }_r*C}$

$Hence,$ ${{(V}_l)}_{rms}$ = $5.75$ $*\frac{1}{50*80\mathrm{}*{10}^{-6}}$ = 1437.5 V

Potential drop across resistor,

${{(V}_R)}_{rms}$ = $I_{rms}$ $*R$ = 5.75 $*40$ = 230 V

Potential drop across LC combination

${{(V}_{LC})}_{rms}$ = $I_{rms}$ $(X_L-X_C)$

At resonance, $X_L$ = $X_C)$ , hence

${{(V}_{LC})}_{rms}=0$

Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.

7.10 Lower tuning frequency, ${\nu }_1$ = 800 kHz = 800 $*{10}^3$ Hz

Upper tuning frequency, ${\nu }_2$ = 1200 kHz = 1200 $*{10}^3$ Hz

Effective inductance of the circuit, L = 200 $\mu H$ = 200 $*{10}^{-6}$ H

Capacitance of variable capacitor for lower tuning frequency ( ${\nu }_1)$ is given as

$C_1$ = $\frac{1}{{\omega }_1^{2}L}$ , where ${\omega }_1$ is the angular frequency for capacitor $C_1$ = 2 $\pi {\nu }_1$

Hence, ${\omega }_1$ = 2 $\pi *$ 800 $*{10}^3$ = 5.026 $*{10}^6$ rad/s

$C_1$ = $\frac{1}{{\omega }_1^{2}L}$ = $\frac{1}{{(5.026*{10}^6)}^2*200\mathrm{}*{10}^{-6}}$ F = 1.9789 $*{10}^{-12}$ F = 197.89 pF

Capacitance of variable capacitor for lower tuning frequency ( ${\nu }_2)$ is given as

$C_2$ = $\frac{1}{{\omega }_2^{2}L}$ , where ${\omega }_2$ is the angular frequency for capacitor $C_2$ = 2 $\pi {\nu }_2$

Hence, ${\omega }_2$ = 2 $\pi *$ 1200 $*{10}^3$ = 7.54 $*{10}^6$ rad/s

$C_1$ = $\frac{1}{{\omega }_2^{2}L}$ = $\frac{1}{{(7.54*{10}^6)}^2*200\mathrm{}*{10}^{-6}}$ F = 87.95 $*{10}^{-12}$ F = 87.95 pF

Hence, the range of variable frequency is from 88 pF to 198 pF.