Practice free Alternating Current (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
When a DC source is connected to a capacitor, what happens to a lamp connected in series with it?
The NCERT example states: 'When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow.'
In an AC circuit, what term is used to describe current that causes no power dissipation?
The NCERT text mentions: 'In a purely inductive or capacitive circuit, $\cos \phi = 0$ and no power is dissipated even though a current is flowing in the circuit. In such cases, current is referred to as a wattless current.'
For a series RLC circuit, resonance occurs when:
The NCERT text states: 'In a RLC circuit, resonance phenomenon occur when $X_L = X_C$ or $\omega_0 = \frac{1}{\sqrt{LC}}$'.
The power factor in an RLC circuit is a measure of:
The NCERT text states: 'The power factor in a RLC circuit is a measure of how close the circuit is to expending the maximum power.'
If a capacitor is connected to an AC source and its capacitance (C) is reduced, what happens to the brightness of a series-connected lamp?
From Example 7.3: 'Reducing C will increase reactance [capacitive reactance $X_C = 1/(\omega C)$] and the lamp will shine less brightly than before.'
According to the NCERT text, what is the relationship between the peak voltage ($v_m$) and peak current ($i_m$) for a purely resistive AC circuit?
From Eq. (7.3) in the NCERT text: '$v_m = i_m R$'. This is essentially Ohm's Law for peak values in a resistive AC circuit.
In a series $LCR$ circuit, the inductance $L$ is $10\ \text{mH}$, capacitance $C$ is $1\ \mu\text{F}$ and resistance $R$ is $100\ \Omega$. The frequency at which resonance occurs is:
$f_0 = \dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{2\pi\sqrt{10^{-2}\times 10^{-6}}} \approx 1.59\ \text{kHz}$.
An ac source is connected to a capacitor $C$. Due to decrease in its operating frequency:
$X_C = 1/(\omega C)$ increases as $\omega$ decreases, so current $I = V/X_C$ (and displacement current) decreases.
A $12\ \text{V}$, $60\ \text{W}$ lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of $220\ \text{V}$. Assuming the transformer to be ideal, what is the current in the primary winding?
Ideal transformer: $V_p I_p = V_s I_s = 60\ \text{W}\Rightarrow I_p = 60/220 \approx 0.27$ A.
The net impedance of circuit (as shown in figure) will be:
$X_L = 2\pi(50)(50/\pi)\times 10^{-3} = 5\ \Omega$. $X_C = 1/[2\pi(50)(10^3/\pi)\times 10^{-6}] = 10\ \Omega$. $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100 + 25} = 5\sqrt{5}\ \Omega$.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.