Class 12 Alternating Current CBSE Questions & Answers
Class 12 · Alternating Current
This is Physics Class 12 Alternating Current CBSE Questions & Answers. There are 15 questions in this test with each question having around four answer choices.
Questions & Answers
1
The correct equation for a series LCR circuit excited by AC is
- A\(L{{{d^2}q} \over {d{t^2}}} + 2R{{dq} \over {dt}} + {q \over C} = 0\)
- B\(L{{{d^2}q} \over {d{t^2}}} + R{{dq} \over {dt}} + {q \over C} = {v_m}sin\omega t\)Correct
- C\(L{{{d^2}q} \over {d{t^2}}} + R{{dq} \over {dt}} + 2{q \over C} = {v_m}sin\omega t\)
- D\(L{{{d^2}q} \over {d{t^2}}} + R{{dq} \over {dt}} + {q \over C} = 0\)
2
The current in a series LCR circuit excited by AC of frequency \(\omega \) is in general
- Aimsin(\(\omega {\rm{t}}\) + 90 )
- B2imsin(\(\omega {\rm{t}}\))
- Cimsin(\(\omega {\rm{t }} + {\rm{ }}\phi \))Correct
- D\({{\rm{i}}_{\rm{m}}}{\rm{sin}}\omega {\rm{t}}\)
3
For a series LCR circuit the input impedance at resonance
- Aequals the resistance RCorrect
- Bequals the resistance \({\rm{R}} + {\rm{j}}\omega {\rm{L}}\)
- Cequals the resistance \(\omega {\rm{L}}\)
- Dequals the resistance \({\rm{1}}/\omega {\rm{C}}\)
4
At resonance the current in an LCR circuit
- Ais zero
- Bis local minimum
- Cis minimum
- Dis maximumCorrect
5
The sharpness of resonance is given by
- A\({\omega _0}{\rm{L}}\) /C
- B\({\omega _0}{\rm{L}}\) C
- C\({\omega _0}{\rm{L}}\) /RCorrect
- D\({\omega _0}{{\rm{L}}^{\rm{2}}}\)/R
6
If V and I are rms voltage and current in an AC circuit, and \({\rm{cos}}\phi \) the power factor the power consumed by a reactive load is in general
- AVI
- BVI \({\rm{sin}}\phi \)
- CVI \({\rm{cos}}\phi \)Correct
- D2VI \({\rm{cos}}\phi \)
7
For a parallel LC circuit the resonant frequency is
- A\({1 \over {LC}}\)
- B\({1 \over {\sqrt {2LC} }}\)
- C\({2 \over {\sqrt {LC} }}\)
- D\({1 \over {\sqrt {LC} }}\)Correct
8
For a parallel ideal LC circuit at resonance the input impedance across L or C is
- Azero
- B\({\omega _0}{\rm{C}}\)
- CinfiniteCorrect
- D\({\omega _0}{\rm{L}}\)
9
Transformer uses the principle of
- AInductionCorrect
- Bleast action
- Ccharge conservation
- Dconduction
10
For a transformer the ratio of output to input equals
- Ainput turns divided by output turns
- Boutput turns divided by twice the input turns
- Coutput turns multiplied by input turns
- Doutput turns divided by input turnsCorrect
11
A light bulb is rated at 50W for a 220 V supply. The resistance of the bulb, the peak voltage of the source and the rms current through the bulb are
- A468 \(\Omega \), 411 V, 0.267A
- B968 \(\Omega \), 350 V, 0.327A
- C768 \(\Omega \), 391 V, 0.297A
- D968 \(\Omega \), 311 V, 0.227ACorrect
12
A pure inductor of 20.0 mH is connected to a source of 230 V. Inductive reactance and rms current in the circuit (if the frequency of the source is 50 Hz). are
- A8.28 \(\Omega \), 36.6 A
- B6.28 \(\Omega \), 36.6 ACorrect
- C6.28 \(\Omega \), 46.6 A
- D9.28 \(\Omega \), 36.6 A
13
A lamp is connected in series with a capacitor and connected to an AC source. As the capacitance value is decreased.
- AThe lamp glows brighter
- BThe lamp starts turning on and off
- CThe lamp does not glow
- DThe lamp glows dimmer and dimmerCorrect
14
25.0 \(\mu {\rm{F}}\) capacitor is connected to a 220 V, 50 Hz source. Capacitive reactance and the current (rms) in the capacitor are circuit
- A127 \(\Omega \), 1.73 ACorrect
- B127 \(\Omega \) , 1.23 A
- C102 \(\Omega \) , 1.03 A
- D111 \(\Omega \) , 1.73 A
15
A light bulb and an open coil inductor are connected to an ac source . On replacing the open coil inductor with an iron core inductor
- AThe bulb glows dimmerCorrect
- BThe bulb glows on and off
- CThe bulb glow is unaffected
- DThe bulb glows brighter