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An alternating voltage $ E=200 \sqrt 2 sin (100t) $ is connected to 1 microfarad capacitor through an ac ammeter. The reading of the ammeter shall be.
In a region of uniform magnetic induction $B = 10^{-2} tesla $, a circular coilof radius 30 cm and resistance $ \pi^ 2 ohm $ is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the coil. If the coil rotates at 200 rpm the amplitude of the alternating current induced in the coil is,
$ I_0 = { NAB \Omega \over R } $
An LCR series circuit with $ R = 100 \Omega $ is connected to a 200 V, 50 Hz a.c source when only the capacitance is removed the current lies the voltage by $ 60 ^ \circ $ when only the inductance is removed, the current leads the voltage by $ 60 ^\circ $ . The current in the circuit is
for inductor $ tan \theta = { \omega L \over R } $ for capacitor $ tan \theta = { 1 \over \omega CR } $ $ |\theta | is same 50 , \omega L = { 1 \over \omega C } $ means resonance $I _ {rms} = { \upsilon _{rms} \over R} $
The power factor of a good choke coil is
The power factor of a choke coil depends on the phase difference between the voltage and the current. For a good choke coil, which is highly inductive, the phase difference is close to 90 degrees. The power factor (
$$ cos \\phi $$
) is given by:
$$ cos \\phi = \\cos (90^\circ) $$
$$ cos \\phi = 0 $$
However, in practical scenarios, it is not exactly zero due to the presence of some resistance. Therefore, it is nearly zero. Thus, the power factor of a good choke coil is nearly zero.
When 100 volt dc is applied across a coil, a current of 1A flows through it. When 100 volt ac at $50 cycle s^{-1} $ is applied to the same coil, only 0.5 A current flows. The impedance of the coil is,
When 100V DC is applied, the resistance (R) of the coil can be found using Ohm's Law: $$ R = rac{V}{I} = rac{100V}{1A} = 100 ext{ ohms} $$ When 100V AC is applied, the current is 0.5A. The impedance (Z) can be found using the same formula: $$ Z = rac{V}{I} = rac{100V}{0.5A} = 200 ext{ ohms} $$ Therefore, the impedance of the coil is 200 ohms.
An LC circuit contains a 20 mH inductor and a $50 \mu F $ capacitor with an initial charge of 10 mc. The resistance of the circuit is negligible. At the instant the circuit is closed be t = 0. At what time is the energy stored compeletly magnetic.
$ t = { T \over 4 } = { 2 \pi \sqrt{LC } \over 4 } $
The quality factor of LCR circuit having resistance (R) and inductance (L) at resonance frequency $( \omega )$ is given by
The quality factor (Q) of an LCR circuit at resonance is given by the formula: $$ Q = \frac{\omega L}{R} $$ where \( \omega \) is the angular frequency, \( L \) is the inductance, and \( R \) is the resistance. This is because, at resonance, the impedance of the circuit is minimized, and the quality factor is defined as the ratio of the energy stored in the circuit to the energy dissipated per cycle.
For high frequency, a capacitor offers
The reactance of a capacitor is given by the formula: $$ X_C = \frac{1}{\omega C} $$ where \( X_C \) is the capacitive reactance, \( \omega \) is the angular frequency, and \( C \) is the capacitance. For high frequencies, the value of \( \omega \) increases, and as a result, the capacitive reactance \( X_C \) decreases. Therefore, a capacitor offers less reactance at high frequencies.
The coil of a choke in a circuit
A choke coil, which is an inductor, opposes changes in current. This property is due to the inductive reactance, which is given by: $$ X_L = \omega L $$ where \( X_L \) is the inductive reactance, \( \omega \) is the angular frequency, and \( L \) is the inductance. As the inductive reactance increases, it resists the flow of current, effectively decreasing the current in the circuit.
The power factor of an ac circuit having resistance (R) and inductance (L) connected in series and an angular velocity w is,
The power factor (PF) of an AC circuit with resistance (R) and inductance (L) connected in series is given by the cosine of the phase angle (Ï•) between the voltage and the current. This phase angle is determined by the impedance of the circuit. The formula for power factor is:
$$ \text{Power Factor} = \cos(\varphi) = \frac{R}{\sqrt{R^2 + (\omega L)^2}} $$
Here, the impedance of the circuit is \( Z = \sqrt{R^2 + (\omega L)^2} \). Thus, the correct option is:
$$ \frac{R}{( R^2 + \omega^2 L^2 )^{1/2}} $$
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.