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# Test: AC Applied Across Inductor

## 10 Questions MCQ Test Physics For JEE | Test: AC Applied Across Inductor

Description
This mock test of Test: AC Applied Across Inductor for JEE helps you for every JEE entrance exam. This contains 10 Multiple Choice Questions for JEE Test: AC Applied Across Inductor (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: AC Applied Across Inductor quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: AC Applied Across Inductor exercise for a better result in the exam. You can find other Test: AC Applied Across Inductor extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### In an a.c. circuit containing inductance only​

Solution:

Solution :- In a purely inductive circuit (an AC circuit containing inductance only) the current lags behind the voltage by π/2

Option is not correct because there is relation between current & voltage in inductance

QUESTION: 2

### L/R has dimensions same as that of​

Solution:

Self-inductance is given by: v(t)=L (di/dt)
From this we have: L=v (dt/di)

v/di has the dimensions of R

So, L=Rdt hence L/R=dt, same units as that of time.

QUESTION: 3

### Q factor is

Solution:

Q-factor: In LCR Circuit, the ratio of resonance frequency to the difference of its neighbouring frequencies so that their corresponding current is 1/√2​ times of the peak value, is called Q-factor of the circuit.
Formula: Q=(1/R)​ √(L/C)​​
So, Q ∝1/R
Conditions for the large value of Q factor:
(i) Value of CL​ should be large.
(ii) Value of R should be less.

QUESTION: 4

If a resistor is connected across the voltage source and the frequency of voltage and current wave form is 50Hz, then what is frequency of instantaneous power

Solution:

P(t)=VmImSin2ωt
P(t)=0.5VmIm(2sin2ωt)
P(t)= 0.5VmIm(1-cos2ωt)
Therefore, frequency is doubled for the instantaneous power so, frequency of instantaneous power is 100Hz.

QUESTION: 5

In inductance there is opposition to the growth of current, it is due to

Solution:

Inductors and chokes are basically coils or loops of wire that are either wound around a hollow tube former (air cored) or wound around some ferromagnetic material (iron cored) to increase their inductive value called inductance.
Inductors store their energy in the form of a magnetic field that is created when a voltage is applied across the terminals of an inductor. The growth of the current flowing through the inductor is not instant but is determined by the inductor's own self-induced or back emf value. Then for an inductor coil, this back emf voltage VL is proportional to the rate of change of the current flowing through it.
This current will continue to rise until it reaches its maximum steady state condition which is around five time constants when this self-induced back emf has decayed to zero. At this point a steady state current is flowing through the coil, no more back emf is induced to oppose the current flow and therefore, the coil acts more like a short circuit allowing maximum current to flow through it.
However, in an alternating current circuit which contains an AC Inductance, the flow of current through an inductor behaves very differently to that of a steady state DC voltage. Now in an AC circuit, the opposition to the current flowing through the coils windings not only depends upon the inductance of the coil but also the frequency of the applied voltage waveform as it varies from its positive to negative values.

QUESTION: 6

When a fluorescent tube is used in A.C. circuit:

Solution:

In a.c. circuits, a choke coil is used to control the current in place of a resistance. If a resistance is used to control the current, the electrical energy will be wasted in the form of heat. Due to its inductive reactance, a choke coil decreases the current without wasting electrical energy in the form of heat.

QUESTION: 7

When Ø is the phase difference, what is the power factor?​

Solution:

Cos Ø is called a power factor that indicates what fraction of [(Voltage V) x (Current I)] becomes the useful power. Most electric circuits have resistance and inductance. In such an electric circuit, the current lags the voltage by the phase difference.

QUESTION: 8

The current and emf through an inductance differ in phase by:​

Solution:
QUESTION: 9

The average power dissipation in pure inductance is:

Solution:

In pure inductance circuit, R=0
Thus power factor of pure inductance circuit cosϕ=R/Z​=0
Average power dissipation Pav​=Vrms​Irms​cosϕ
⟹ Pav​=0

QUESTION: 10

In purely inductive circuits, the current:

Solution: