R must be short-circuited in order for the circuit to become an LC oscillator. The capacitor will continue to drain, and all of the energy will be transferred to L and back. Energy will oscillate...
, the voltage source stops and R is short-circuited. Now write down how much energy is stored in each of L and C.i) The equation for charge motion variation with respect to time is as follows: $\mathrm{L} d^{2} q(\mathrm{t}) / \mathrm{dt}+\mathrm{R} \mathrm{dq}(\mathrm{t}) / \mathrm{dt}+\mathrm{q}(\mathrm{t})...
i) The first equation is in the form of conservation of energy ii) We get dt as positive when we integrate the equation from $0$ to $T$, which is possible when the phase difference is constant and...
Given equation is, $L d i / d t+R i+q / C=v i=v m \sin \omega t$ i) On multiplying the above equation with I, we get, $\mathrm{d}\left(1 / 2 \mathrm{Li}^{2}\right) / \mathrm{dt}+1 / 2 \mathrm{C}...
The impedance $Z$ for the given circuit is represented as, $1 / \mathrm{Z}=\left[1 / \mathrm{R}^{2}+1 /(1 / \omega \mathrm{C}-\omega \mathrm{L})^{2}\right]^{1 / 2}$
i) It is given that the power station is $10 \mathrm{~km}$ away from the town Length of the $\mathrm{Cu}$ wire is given as $\mathrm{L}=20 \mathrm{~km}=20000 \mathrm{~m}$ Resistance of $\mathrm{Cu}$...
,
. Another device has twice the values for 
, 
and 
. How are the answers affected?Impedance is given as Z = 25 ohms $635=25R^{2}/16$ a) Resistance can be calculated as, $R=\sqrt 25\times 16=\sqrt 400=20 ohms$ b) $X_{c}-X_{l}=-3R/4=-15ohms$ c) Main current will be...
The reactance offered by an inductor increases with the increasing frequency of an alternating voltage because the induced emf is proportional to the rate of change of current.
Because capacitance reactance is inversely proportional to frequency, and capacitors have a tendency to pass high-frequency current while blocking low-frequency currents, the reactance delivered by...
$P_{s}$ is given as $60W$ $I_{s}$ is given as $0.54A$ Primary voltage is given as $220V$ $V_{s}=60/0.54=110V$ The transformer is a step-down transformer because the secondary voltage is lower than...
Inductance L is given as 0.01 H Resistance R is given as 1 ohm Voltage is given as 200 V Frequency is given as 50 Hz Impedance is given Z = 3.3 ohms Phase difference will be 72 × π/180 rad 0.004 sec...
The phenomenon called Joule's law of heating is used to determine the ampere for an alternating current. According to Joule's law of heating in a one-ohm resistance, AC is the current produced in a...
Solution: X could be an inductor, a capacitor, or a hybrid of the two.
Solution: a) The greatest amplitude of the power curve is equal to the product of the voltage and current amplitudes. As a result, curve A denotes power. b) Because the whole cycle in the graph...
$tan \phi = \frac{X_{l}-X_{c}}{R}$ $X_{l}<X_{c}$ At resonate frequency we have, $X_{l}=X_{c}$ $tan \phi = 0$
Solution: $I_{rms}=1.6A$
Frequency 1 is 0.8 rad/s Frequency 2 is 1.2 rad/s The bandwidth is 0.4 rad/s
a) When P < 0, then AC source's instantaneous power output can be negative. b) The average power output cannot be negative as $P_{av}>0$.
(a) Under which conditions would the rms currents in the two circuits be the same? (b) Can the rms current in the circuit (b) be larger than that in (a)? Solution: a) $(I_{rms})a=(I_{rms})b$...
Solution: The effective impedance will be, $Z_{eq}=R_{1}+R_{3}$
The magnetic energy of the LC circuit is equivalent to kinetic energy, while electrostatic energy due to a capacitor's charge change is analogous to potential energy.
The correct options are: (a) zero average current. (d) voltage and current possibly differing in phase φ such that |φ| < π/2.
The correct options are: (c) the charge on the plates is in phase with the applied voltage. (d) power delivered to the capacitor is zero.
The correct options are: (a) Here, the power factor cos 0, 0. φ ≥ ≥ P (b) The driving force can give no energy to the oscillator (P = 0) in some cases. (c) The driving force cannot syphon out (P...
The correct options are: (a) For a given power level, there is a lower current. (b) Lower current implies less power loss. (d) It is easy to reduce the voltage at the receiving end using step-down...
The correct options are: (c) Resistor and a capacitor. (d) Only a capacitor.
The correct options are: (a) Inductor and capacitor. (d) Resistor, inductor and capacitor.
(a) 1/ √2 A.
(c) 14.4 W.
(c) R = 15 Ω, L = 3.5 H, C = 30µF.
(b) another capacitor should be added in parallel to the first
and is calibrated to read 
.(c) the meter reads not v but $<v_{2}>$ and is calibrated to read $\sqrt <v_{2}>$
and an internal reactance 
. It is used to supply power to a passive load consisting of a resistance and a reactance . For maximum power to be delivered from the generator to the load, the value of is equal to.
..(c) $-X_{g}$.
(b) 5 √3/2 A
Capacitance of the capacitor, $\mathrm{C}=100 \mu \mathrm{F}=100 \times 10^{-6} \mathrm{~F}$ The resistance of the resistor, $\mathrm{R}=40 \Omega$ Supply voltage, $\mathrm{V}=110 \mathrm{~V}$...
Capacitance of the capacitor, $\mathrm{C}=100 \mu \mathrm{F}=100 \times 10^{-6} \mathrm{~F}$ The resistance of the resistor $=40 \Omega$ Supply voltage, $\mathrm{V}=110 \mathrm{~V}$ Frequency, $v=60...
Therefore, RMS current in the line, $1=\left(800 \times 10^{3}\right) / 40,000$ $$\mathrm{I}=20 \mathrm{~A}$$ Total resistance of the two wire line, $\mathrm{R}=2 \times 15 \times 0.5=15 \Omega$ (a)...
Power required, $P=800 \mathrm{~kW}=800 \times 10^{3} \mathrm{~W}$ Power = Voltage $x$ current $\Rightarrow 800 \times 10^{3}=4000 \times 1$ Therefore, RMS current in the line, $I=\left(800 \times...
$$\mathrm{h}=300 \mathrm{~m}$$ Flowing water volume $=100 \mathrm{~m}^{3} \mathrm{~s}^{-1}$ Density of water, $\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}$ Electric power, $\mathrm{P}=\mathrm{h} \rho...
The transformer's input voltage, $\mathrm{V}_{1}=2300 \mathrm{~V}$ The step-down transformer's primary windings, $\mathrm{n}_{1}=4000$ turns Output voltage, $V_{2}=230 \mathrm{~V}$ Let be the...
(a) Of course. The instantaneous voltage applied is equal to the algebraic sum of the instantaneous voltages across the circuit's series parts. The same cannot be said for rms voltage since voltages...
Inductance, $L=3.0 \mathrm{H}$ Capacitance, $\mathrm{C}=27 \mu \mathrm{F}=27 \times 10^{-6} \mathrm{~F}$ Resistance, $\mathrm{R}=7.4 \Omega$ The resonant frequency of the source in the LCR series...
The LC oscillations are eventually dampened by the resistor. The entire original energy (= 1.0 J) is dissipated as heat in the end.
ATQ, the frequency range of (ν) of radio is $800to1200kHz$ ATQ, the Lower tuning frequency of the circuit would be, ${{v}_{1}}=800kHz=800\times {{10}^{3}}Hz$ ATQ, the Upper tuning frequency of the...
At resonance in the circuit, the supply frequency and the natural frequency are equal. $R=20\Omega $ is the resistor's resistance. Given the inductor's inductance, $L=1.5H$ Given the...
According to the question, Capacitance value of the capacitor will be, $C=30\mu F=30\times {{10}^{-6}}F$ Inductance of the inductor will be, $L=27mH=27\times {{10}^{-3}}H$ Charge on the...
Given Capacitance value of the capacitor is , $C=30\mu F=30\times {{10}^{-6}}F$ Given Inductance value of the charged inductor will be , $L=27mH=27\times {{10}^{-3}}H$ Angular frequency according to...
The inductance of the inductor can be considered as $L=2.0H$ The capacitance of the capacitor will be , $C=32\mu F=32\times {{10}^{-6}}F$ The resistance of the resistor as per question is...
The capacitance of the capacitor in the circuit is given as $C=60\mu For60\times {{10}^{-6}}F$ The source voltage is considered to be $V=110V$ The frequency of the source will be $v=60Hz$ The...
As per given : The inductor connected to the given AC supply has an inductance $L=44mH=44\times {{10}^{-3}}H$ The magnitude of the source voltage V is considered to be $220V$ The frequency of the...
a)The peak voltage of the AC supply will be ${{V}_{o}}=300V$ And we also know that, ${{V}_{RMS}}=V=\frac{{{V}_{o}}}{\sqrt{2}}$ Substituting the values, we get to know that...
(a) What is the rms value of current in the given circuit? (b) What will be the net power consumed over a full cycle? Assume: The resistor's resistance R is 100. 220 V is the source voltage. The...