Alternating Current Online Objective Test | Alternating Current Online Objective Test

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q1.

If Vm and Im are peak voltage and current, Impedance Z in an AC circuit is

A. vmim

B. 2vmim

C. vm2im

D. vmim

Answer : Option D
Explaination / Solution:

Impedance refers to the overall obstruction offered by a circuit containing different components to the passage of current. Also Z has unit same as resistance.

So, drawing an analogy with the ohm's law, we get the above relation

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q2. The correct equation for a series LCR circuit excited by AC is

A. Ld2qdt2+Rdqdt+qC=vmsinωt

B. Ld2qdt2+Rdqdt+qC=0

C. Ld2qdt2+2Rdqdt+qC=0

D. Ld2qdt2+Rdqdt+2qC=vmsinωt

Answer : Option A
Explaination / Solution:

The LHS of equation contains expression for current through inductor, resistor and capacitor respectively. Each expression on LHS is a result of basic mathematical derivation. On the rhs we have the overall current in the circuit

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q3. The current in a series LCR circuit excited by AC of frequency ω is in general

A. Imsinωt

B. Imsin(

ω t

+ 90 )

C. 2Imsin(

ω t

)

D. Imsin (ωt+ϕ)

Answer : Option D
Explaination / Solution:

The above expression is for instantaneous value of current in an AC Circuit and since in an LCR circuit current is same across the circuit, the above expression holds valid for the entire LCR series circuit.

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q4. For a series LCR circuit the input impedance at resonance

A. equals the resistance ωL

B. equals the resistance R+jωL

C. equals the resistance R

D. equals the resistance 1/ωC

Answer : Option C
Explaination / Solution:

Impedance in series LCR circuit

$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$

at resonance

$\begin{aligned} X_{_{L}} = X_{C} \\ ω L = \frac{1}{ω C} \end{aligned}$

$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}} = R$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q5. At resonance the current in an LCR circuit

A. is minimum

B. is zero

C. is local minimum

D. is maximum

Answer : Option D
Explaination / Solution:

in LCR series circuit current

$i = \frac{E}{Z} = \frac{E}{\sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}$

at resonance

$\begin{aligned} X_{_{L}} = X_{C} \\ ω L = \frac{1}{ω C} \\ Z_{min} = R \end{aligned}$

hence current will be maximum.

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q6. The sharpness of resonance is given by

ω_{0} L

ω_{0} L

ω_{0} L

ω_{0} L^{2}

Answer : Option B
Explaination / Solution:

Sharpness of resonance is quantitatively described by a dimensionless number known as Q-factor or quality factor which is numerically equal to ratio of resonant frequency to bandwidth.

therefore

Q-factor = resonant frequency / bandwidth

sharpness of resonance = $Q = \frac{ω_{0}}{Δ ω} = \frac{ω_{0}}{R / L} = \frac{ω_{0} L}{R}$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q7. For a parallel ideal LC circuit at resonance the input impedance across L or C is______

A. zero

LC√LC√

B. Equal to resistance

C. equal to XL

D. infinite

Answer : Option D
Explaination / Solution:

in parallel LC circuit, Impedance

$\begin{aligned} Z = \frac{ω L}{ω^{2} L C - 1} = (\frac{1}{C}) (\frac{ω}{ω^{2} - {ω_{0}}^{2}}) \\ (ω_{0} = \frac{1}{\sqrt{L C}}) \end{aligned}$

at resonance $ω = ω_{0}$

hence impedance will be infinite.

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q8. Transformer uses the follwing principle of

A. least action

B. charge conservation

C. Induction

D. conduction

Answer : Option C
Explaination / Solution:

A transformer consists of two electrically isolated coils. It works on the principle of mutual induction of two coils or Faraday Law’s Of Electromagnetic induction. When current in the primary coil is changed the flux linked to the secondary coil also changes.hence an EMF is induced in the secondary coil due to Faraday law’s of electromagnetic induction.

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q9. 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

A. 9.28 Ω, 36.6 A

B. 6.28 Ω, 46.6 A

C. 6.28 Ω, 36.6 A

D. 8.28 Ω, 36.6 A

Answer : Option C
Explaination / Solution:

$\begin{aligned} L = 20 m H = 20 \times 10^{- 3} H \\ V = 230 v o l t \\ f = 50 H z \end{aligned}$

inductive reactance

$X_{L} = ω L = 2 π f \times L = 2 \times 3.14 \times 50 \times 20 \times 10^{- 3} = 6.28 Ω$

rms current

$i = \frac{V}{X_{L}} = \frac{230}{6.28} = 36.6 A$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q10. 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

A. 768 Ω, 391 V, 0.297A

B. 968 Ω, 311 V, 0.227A

C. 468 Ω, 411 V, 0.267A

D. 968 Ω, 350 V, 0.327A

Answer : Option B
Explaination / Solution:

$\begin{aligned} P = \frac{V^{2}}{R} \\ 50 = \frac{220 \times 220}{R} \\ R = \frac{220 \times 220}{50} = 968 Ω \end{aligned}$

Peak value of voltage

$\begin{aligned} V_{0} = V_{r m s} \sqrt{2} \\ V_{0} = 220 \times \sqrt{2} = 311 V \end{aligned}$

rms value of current

$i_{r m s} = \frac{V_{r m s}}{R} = \frac{220}{968} = 0.227 A$

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