Circuit Elements MCQ Quiz in मल्याळम - Objective Question with Answer for Circuit Elements - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Mathematically, Resistance is defined as:

\(R=\rho \frac{L}{A}\)

R is the resistance of the wire

A is the area of cross-section of the wire

L is the length of the wire.

Current division rule

By using the current division rule,

The current flows through the resistor R­1 = I1 = \(\frac{{I\left( {{R_2}} \right)}}{{{R_1} + {R_2}}}\) .......(1)

The current flows through the resistor R­2 = I2 = \(\frac{{I\left( {{R_1}} \right)}}{{{R_1} + {R_2}}}\)......... (2)

From equation (1) & (2) 

\(\frac{I_1}{I_2}=\frac{R_2}{R_1}\) ......... (3)

Calculation:

Given

Current I₁ = 80 % more than I₂ = 1.8 I₂

Length L₂ = 45 % longer than L₁ = 1.45 L₁

ρ₁ : ρ₂ = 40 : 20  

From equation (3) 

\(\frac{R_2}{R_1}=\frac{I_1}{I_2}\Rightarrow \frac{R_2}{R_1}=\frac{1.8I_2}{I_2}=1.8\)

As resistance is given as \(R=\rho \frac{L}{A}\)

So \(\frac{R_2}{R_1}=\frac{\rho_2 \frac{L_2}{A_2}}{\rho_1 \frac{L_1}{A_1}}= \frac{\rho_2L_2A_1}{\rho_1L_1A_2}\)

\(\Rightarrow 1.8 =\frac{\rho_2L_2A_1}{\rho_1L_1A_2}=\frac{20\;\times\;1.45L_1A_1}{40\;\times\;L_1A_2}\)

\(\Rightarrow \frac{A_2}{A_1}=0.402\)

Conductivity

The specific conductance of conductivity is the reciprocal of the resistivity.

\(σ={1\over ρ}\)

The resistance of a conductor is given by:

\(R={ρ l\over A}\)

\({1\over \rho}={l \over RA}\)

\(\sigma={l \over RA}\)

where, σ = Conductivity

ρ = Resistivity

l = Length of conductor

A = Cross-sectional area of the conductor

The unit of conductivity is given by:

\(\sigma ={S\times m\over m^2}\)

\(\sigma ={S\over m}\)

∴ The unit of conductivity is Siemens/Metre.

Note: The unit of Resistivity is an Ohm-Meter

Concept:

The resistance of a conductor is given by:

\(R = \rho\frac{{ ~l}}{A}\)

ρ = resistivity

R = Resistance

l = length of wire

A = cross-sectional area of the wire

Application:

Let R1 = Resistance of 1st copper conductor

R2 = Resistance of 2nd copper conductor

\(\frac{R_1}{R_2}=\frac{l_1/A_1}{l_2/A_2}\)

Since A = πr2, the above can be written as:

\(\frac{R_1}{R_2}=\frac{l_1/\pi r_1^2}{l_2/\pi r_2^2}=\frac{l_1r_2^2}{l_2r_1^2}\)

Given \(\frac{l_1}{l_2}=\frac{1}{4}\)and \(\frac{r_1}{r_2}=\frac{1}{2}\)

\(\frac{R_1}{R_2}=\frac{1}{4}\times (\frac{2}{1})^2\)

\(\frac{R_1}{R_2}=\frac{1}{1}\)

Concept:

Self-inductance \(L = \frac{{{{\rm{μ }}_0}{{\rm{μ }}_{\rm{r}}}{\rm{A}}{{\rm{N}}^2}}}{l}\) ... (1)

Here, A = Area

N = Number of turns

l = Length of coil

Reluctance (S) = \(\frac{1}{\mu_{0}\mu_{r}}\times \frac{l}{A}\) ... (2)

From equation (1) and (2)

Reluctance \(s = \frac{{{N^2}}}{L}\)

Calculation:

L = 4 H, s = 100 AT / Wb

From equation (i)

\(N = \sqrt {4 \times 100} \)

= 20

CONCEPT:

Self-Induction

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
• Self-inductance of a solenoid is given by –

\(⇒ L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)

Where μo = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = Area of the solenoid.

EXPLANATION:

Given - l2 = 2l1

N₂ = 2 N₁

Self-inductance of a solenoid is given by:

\(⇒ L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)

\(⇒ L\propto \frac{N^2}{l}\)

\(\Rightarrow \frac{L_2}{L_1}=\frac{N_2^2}{N_1^2}\times \frac{l_1}{l_2}=\frac{(2N_1)^2}{N_1^2}\times \frac{l_1}{2l_1}=2\)

L₂ = 2 L_1​

Concept:

A capacitor is a device used to store energy.

The process of charging up a capacitor involves the transferring of electric charges from one plate to another.

The work done in charging the capacitor is stored as its electrical potential energy.

The energy stored in the capacitor is

\(U = \;\frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)

Where,

Q = charge stored on the capacitor

U = energy stored in the capacitor

C = capacitance of the capacitor

V = Electric potential difference

Calculation:

Capacitance (C) = 100 μF  = 100 × 10-6 F

Applied voltage V₁= 100 V and V₂ = 500 V

The energy stored in the capacitor is

= \(\frac{1}{2} \times 100\times 10^{-6} \times \left( {V_{2\;}^2 - V_1^2} \right)\)

= \(\frac{1}{2} \times 100\times 10^{-6} \times \left( {{{500}^2} - {{100}^2}} \right)\)

= 12 J

Concept:

The below table shows the colour coding and their respective tolerances for the calculation of resistance:

Steps for calculation of resistance:

• The last colour to the right represents the tolerance in 5 colour bands whereas it represents the TCR in 6 band colours.
• The second last colour from the left represents the multiplier.
• Starting three colours represents the value of resistance.

Calculation:

From the given colour band the silver represents the tolerance (± 10%).

The Brown represents the multiplier (10).

The starting three colours represent the value.

Orange = 3, Yellow = 4, and Green = 5.

Total resistance is:

R = 345 × 10¹ ± 10%

Option 2 is correct.

Concept:

Energy stored in condenser:

• A capacitor is a device that stores energy in the form of a charge.
• The process of charging up a capacitor involves the transferring of electric charges from one plate to another.
• The work done in charging the capacitor is stored as its electrical potential energy.

The energy stored in the capacitor is:

\(U = \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)

Where, 

Q = charge stored on the capacitor 

U = energy stored in the capacitor 

C = capacitance of the capacitor

V = Electric potential difference

Calculation: 

Given:

C = 50 μF, V = 200V

The energy stored in the capacitor is:

\(U = \frac{1}{2}C{V^2} \)

\(U = \frac{1}{2}(50\times10^{-6})\times(200){^2} \)

U = 1J

Concept

The specific conductance of conductivity is the reciprocal of the resistivity.

\(σ={1\over ρ}\)

The resistance of a conductor is given by:

\(R={ρ l\over A}\)

\({1\over \rho}={l \over RA}\)

∵ \({1\over R}=G\)

\(\sigma={Gl \over A}\)

where, σ = Conductivity

ρ = Resistivity

G = Conductance

l = Length of conductor

A = Cross-sectional area of the conductor

The unit of conductivity is given by:

\(\sigma ={S\times m\over m^2}\)

\(\sigma ={S\over m}\)

∴ The specific conductance of a conductor is given by σ = \(\rm\frac{Gl}{A}\) S/m

Calculation:

Since the network contains infinite resistance, we can redraw the network as:

Now equating the equivalent resistance with RAB, we get:

\(\therefore {R_{AB}} = \frac{{{2 R_{AB}}}}{{{R_{AB}} \ + \ 2}} + 1\)

R2AB + 2RAB = 3 RAB + 2

R2­AB - RAB - 2 = 0

Now solving the above quadratic equation for RAB, we get

R2­AB - 2RAB + R_AB - 2 = 0

RAB(RAB - 2) + 1(RAB - 2) = 0

RAB = 2 and RAB = -1

RAB = 2 Ω (∵ Resistance cannot be negative)

Hence option (2) is the correct answer.