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Answer the following questions.
(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a de voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across $C$ and the ac signal across $L$.
(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp's brightness. Predict the corresponding observations if the connection is to an ac line.
(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a de voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across $C$ and the ac signal across $L$.
(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp's brightness. Predict the corresponding observations if the connection is to an ac line.
(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
Solution:
1372 Upvotes
Verified Answer
(a) It is true that applied instantaneous voltage is equal to algebraic sum of instantaneous potential drop across each circuit element in series.
$$
\begin{aligned}
&E=V_R+V_L+V_C \\
&E_0 \sin \omega \mathrm{t}=\frac{E_0}{R} \sin \omega t \\
&\quad+\frac{E_0}{X_L} \sin \left(\omega t-\frac{\pi}{2}\right)+\frac{E_0}{X_C} \sin \left(\omega t+\frac{\pi}{2}\right)
\end{aligned}
$$
But the rms voltage applied is equal to vector sum of potential drop across each element, as voltage drops are in different phase.
(b) At the time of broken circuit of the induction coil the high voltage induced charges the capacitor. This avoid sparking in the circuit.
(c) Inductive reactance $X_{\mathrm{L}}=2 \pi f l$
For a.c. $X_C \propto f$ For d.c. $f=0, X_L=0$
Capacitive reactance $X_C=\frac{1}{2 \pi f C}$
For a.c. $X \propto \frac{1}{f}$
For d.c. $f=0 X_{\mathrm{C}}=\infty$
So, superimpose applied voltage will have all d.c potential drop across $X_{\mathrm{C}}$ and will have most of a.c potential drop across $X_{\mathrm{L}}$.
(d) Inductor offer no hinderance to d.c $X_L=0$, so insertion of iron core does not effect the d.c. current or brightness of lamp connected.
But it definitely effect a.c. current as inserted of iron core increases ' $L$ ' $L=\mu_{\mathrm{m}} n I$
thus increase $X_{\mathrm{L}}(2 \pi f$ L). a.c. current in the circuit reduce $I_{\mathrm{v}}=\frac{E_v}{X_L}$ and brightness of the bulb also reduce.
(e) A fluorescent tube is connected directly across a $220 \mathrm{~V}$ source, it would draw large current which may damage the filaments of the tube, so a choke coil which behaves as $L-R$ circuit reduces the current to appropriate value, and that also which a lesser power loss.
$$
I_v=\frac{E_v}{\sqrt{R^2+X_L^2}}, \quad \mathrm{P}=E_{\mathrm{v}} I_{\mathrm{v}} \cos \phi
$$
An ordinary resistor used to control the current would have maximum power wastage as heat
$$
I_v=\frac{E_v}{R}, P_{\max }=E_{\mathrm{v}} I_{\mathrm{v}}
$$
$$
\begin{aligned}
&E=V_R+V_L+V_C \\
&E_0 \sin \omega \mathrm{t}=\frac{E_0}{R} \sin \omega t \\
&\quad+\frac{E_0}{X_L} \sin \left(\omega t-\frac{\pi}{2}\right)+\frac{E_0}{X_C} \sin \left(\omega t+\frac{\pi}{2}\right)
\end{aligned}
$$
But the rms voltage applied is equal to vector sum of potential drop across each element, as voltage drops are in different phase.
(b) At the time of broken circuit of the induction coil the high voltage induced charges the capacitor. This avoid sparking in the circuit.
(c) Inductive reactance $X_{\mathrm{L}}=2 \pi f l$
For a.c. $X_C \propto f$ For d.c. $f=0, X_L=0$
Capacitive reactance $X_C=\frac{1}{2 \pi f C}$
For a.c. $X \propto \frac{1}{f}$
For d.c. $f=0 X_{\mathrm{C}}=\infty$
So, superimpose applied voltage will have all d.c potential drop across $X_{\mathrm{C}}$ and will have most of a.c potential drop across $X_{\mathrm{L}}$.
(d) Inductor offer no hinderance to d.c $X_L=0$, so insertion of iron core does not effect the d.c. current or brightness of lamp connected.
But it definitely effect a.c. current as inserted of iron core increases ' $L$ ' $L=\mu_{\mathrm{m}} n I$
thus increase $X_{\mathrm{L}}(2 \pi f$ L). a.c. current in the circuit reduce $I_{\mathrm{v}}=\frac{E_v}{X_L}$ and brightness of the bulb also reduce.
(e) A fluorescent tube is connected directly across a $220 \mathrm{~V}$ source, it would draw large current which may damage the filaments of the tube, so a choke coil which behaves as $L-R$ circuit reduces the current to appropriate value, and that also which a lesser power loss.
$$
I_v=\frac{E_v}{\sqrt{R^2+X_L^2}}, \quad \mathrm{P}=E_{\mathrm{v}} I_{\mathrm{v}} \cos \phi
$$
An ordinary resistor used to control the current would have maximum power wastage as heat
$$
I_v=\frac{E_v}{R}, P_{\max }=E_{\mathrm{v}} I_{\mathrm{v}}
$$
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