Inductors and Inductance MCQ Quiz in தமிழ் - Objective Question with Answer for Inductors and Inductance - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Self-inductance and Inductor:

• Self Inductance is the property of a coil by which an emf is induced in it by changing the associated magnetic flux
• The change in magnetic flux is depicted by the change in current in the coil.
• The unit of self-inductance is Henery.

\( e=L \frac{dI}{dt} \)

e is emf induced, L is self-inductance, \(\frac{di}{dt}\) is rate of change in current with time.

• The electrical device with specified Inductance is called Inductor.
• The energy in Inductor: The energy of the inductor is given as 

\(U = \frac{1}{2} LI^2\)

L is inductance, I is current.

Calculation:

Given Inductance L = 100 mH = 100 × 10 ^-3 H = 10 ^-1 H

Current I = 1 A

Energy \(U = \frac{1}{2} 10^{-1}(1)^2 J\) 

\(\implies U = \frac{1}{20}J\)

⇒ U = 0.05 Joule

So, Energy is 0.05 Joule.

0.05 J is the correct option.

CONCEPT:

• An inductor coil is used to suppress or limit the flow of alternating current without affecting the flow of direct current.
  • As in applications that require conversions of AC to DC.
• When current flows through the inductor, it starts storing energy in the form of a magnetic field 

EXPLANATION :

• The power stored in an inductor is given by

⇒ P = Vi      -----(1)

Where V = Voltage induced in an inductor, and i = current 

• The voltage induced in the inductor is given by

\(⇒V =L\frac{dI}{dt}\)

Substituting the above equation in equation 1, it becomes 

\(⇒P = L i \frac{dI}{dt}\)

• The energy stored in the inductor is given by

\(⇒E = \int P dt\)

Substituting the value of P in the above equation

\(⇒ E = \int L i\frac{dI}{dt}dt\)

\(⇒ E = \int L i di\)

\(⇒ E = \frac{1}{2} LI^{2}\)

• The above equation gives the energy stored in the inductor. Hence, option 2 is the answer

Concept:

• The potential difference across inductors: 
  •  V_L = -L dI/ dt
• The potential difference across resistors:
  • V_R = I R

Calculation:

            V_A- (5 X 1) -15 - 5 X 10^-3 X 103 = V_B

             V_A-5-15-0.515 = V_B

             V_A- V_B = 20.515

Conclusion:

The VA – VB  value is 20 V.

Concept:

The power dissipated in the RC circuit depends on both the resistor and the capacitor.

The impedance for the RC circuit is given by:

Z_RC = [(R₂ + (1/ωC)²)]^1/2

The power factor for an RC circuit is: cos(φRC) = R / Z_RC

​Explanation:

• The power dissipated in the RC circuit is:
  • P1 = (V²_rmsR) / Z²RC
  • The power dissipated in the RL circuit depends on both the resistor and the inductor.
• The impedance for the RL circuit is given by:
  • Z_RL = [(R₂ + (ωL)²)]^1/2
  • The power factor for an RL circuit is:
  • cos(φRL) = R / Z_RL
• The power dissipated in the RL circuit is:
  • P2 = (V²_rms R) / Z²_RL
• In an LC circuit, the inductive reactance and capacitive reactance cancel each other out when
  • ωL = 1/ωC.
• This results in a purely reactive circuit with no real power dissipation. Hence, the power in the LC circuit is:
  • P3 = 0

Thus, Both the RC and RL circuits dissipate real power, and because both involve resistive elements, they will dissipate the same amount of power.
The LC circuit, being purely reactive, dissipates no real power.
P1 = P2 > P3 

∴ The correct option is 3

CONCEPT:

• Overall Impedance (Z) in a series combination of a resistor (R), a capacitor (C):

\(Z = \sqrt{R^2 + \frac{1}{ω C}^2}\)

where R is resistance and 1/ωC is Capacitance reactance.

• Overall Impedance (Z) in a series combination of a resistor (R), an inductance (L):

\(Z = \sqrt{R^2 + {ω L}^2}\)

where R is resistance and ωL is Inductance reactance.

The self-inductance of a coil is given by:

\(L =N { ϕ \over I}\)

where L is the self-inductance, N is the number of turns, ϕ is the number of flux, and I is the current in the coil.

For the parallel plate capacitor, the capacitance is given by:

\(C= \frac{KA\varepsilon _{0}}{d}\)

Where A is the area of the plate, d is the distance between the plate, K is the dielectric constant of material and ϵ is the constant.

• The brightness of the bulb is directly proportional to the current passing through it.
• The brightness of the bulb increases, when the effective resistance of the bulb decreases and so current in the bulb increases.

EXPLANATION:

Let the resistance of the bulb is R

• Option 1: If we increase the separation between the plates of the capacitor, the capacitance will decrease. So R_eff​ will increase which will decrease the current. Hence the brightness of the bulb will decrease.
• Option 2: The dielectric will increase the capacitance. So R_eff​ will decrease, so current will increase and thus the brightness of the bulb will increase. So this is the correct option.
• Options 3 and 4: Both cases will increase the inductance of the coil making R_eff​ higher, which will reduce the current and makes the bulb less bright.
• Thus the correct answer is option 2.

Concept: 

The current LR circuit is given by 

 \(I_{rms}={V_{rms} \over \sqrt{R^2 + ω ^2L^2}}\)

where V is Velocity, R is Resistance ω is the frequency and L is Inductance.

Solution: 

We know Current 

\(I_{rms}={V_{rms} \over \sqrt{R^2 + ω ^2L^2}}\)

⇒\(10 = {160 \over \sqrt {25 + (2π 50L)^2}}\)

⇒\(100 = {160^2 \over 25+(2 π 50 L)^2 }\)

⇒\(25+(100\pi L)^2 =256\)

⇒L = 0.048 = 4.84 ×10^-2 H

The correct answer is option (1)

The correct answer is option no "3"

Concept:-

The power dissipated as heat is written as;

 P = i2R          

Here we have i as the current and R as the resistance and P as the power dissipated.

The time constant of the circuit is written as;

\(τ = \frac{L}{R}\)           

Here, \(τ\) is the time constant and L is the inductance and R is the resistance.

Energy stored in an inductor is written as;

U = \(\frac{1}{2}Li^2\)         
Here, L is the inductance, i is the current and U is the energy stored.

CALCULATION:

The power dissipated as the heat is written as;

P = i2R

⇒ 10 = \((1)^2\) × R

⇒ R = 10 Ω 

Now, the time constant of the circuit is written as;

\(τ = \frac{L}{R}\)     ----(1)

Now, on putting the values of R in equation (1) we have;

⇒ \(10~\times 10^-3~= \frac{L}{10}\)

L= 100 mH

Concept:

When two coils are brought near to each other, then the magnetic flux linked with one coil leads to generation of voltage in the second coil due to changes in its magnetic flux, this property is known as mutual induction.

Because of Mutual induction voltage (Current) of a coils changes due to magnetic field of another coil brought near to it.

Calculation:

Two concentric coils are of radius R1 and R2 as shown

Let current in outer loop be i

Magnetic field at centre, \( B= \frac{{\mu_0 i}}{{2{R_1}}}\)

Magnetic flux through inner coil \( = B \times \pi R_2^2\)

\(ϕ = \frac{{{\mu _0}i}}{{2{R_1}}} \times \pi R_2^2\)

\(ϕ = \frac{{{\mu _0}i}}{2} \times \frac{{\pi R_2^2}}{{{R_1}}}\)

as per definition, ϕ = Mi

\( \Rightarrow M = \left( {\frac{{{\mu _0}\pi }}{2}} \right)\frac{{R_2^2}}{{{R_1}}}\)

\(\therefore M \propto \frac{{R_2^2}}{{{R_1}}}\)

CONCEPT:

• Inductor: A device or component of a circuit that has significant self-inductance and stores energy in a magnetic field.
• In the Series circuit, the equivalent Inductance is the algebraic sum of all the inductances.

Leq = L1 + L2 + L3 +......  (In Series)

• And in the Parallel circuit, the reciprocal of the equivalent inductances is the algebraic sum of all the reciprocal of the inductances.

1/Leq = 1/L1 + 1/L2 + 1/L3 +...... (in Parallel)

EXPLANATION:

• In the question, the given circuit has two series-connected inductors.

So equivalent inductance of the circuit will be

Leq = L1 + L2 

Hence the correct answer is option 2.

CONCEPT:

The impedence of the circuit is written as;

\(Z = X_c-X_L\)

Here \(X_c = \frac{1}{\omega C}\) and \(X_L = \omega L\)

CALCULATION:

Given:  L = 40 mH 

Capacitance C = 100 μF

Angular frequency, \(\omega = 314\) rad/s

As we know;

\(Z =\frac{1}{\omega C}- \omega L\)

⇒ \(Z =\frac{1}{314 \times 100 \times 10^{-6}}-314 \times 40 \times 10^{-3}\)

⇒ \(Z =19.28 \Omega\)

Now, \(X_C >X_L\) so the current is leading by \(\frac{\pi}{2}\)

The current is written as;

\(i= \frac{V_o}{Z}sin(\omega t +\frac{\pi}{2})\)

⇒ \(i= \frac{10}{19.28}cos(\omega t )\)

⇒ \(i=0.52 cos(314t)\)

Hence option 2) is the correct answer.