Resistance MCQ Quiz - Objective Question with Answer for Resistance - Download Free PDF

Concept:

Wire wound resistors are a type of precision resistor in which resistance is provided by a metal wire wound around an insulating core. The wire material must have high resistivity and stability over a wide range of temperatures.

Calculation

The most commonly used material for winding in wire wound resistors is a nickel-chromium alloy (such as Nichrome). This alloy offers high electrical resistance, good thermal stability, corrosion resistance, and mechanical strength, making it ideal for resistor applications.

Other materials like carbon are used in carbon composition resistors, while chromium cobalt and pure nickel are not commonly used for wire wound resistor windings.

Concept

The below equation shows the method to find the resistance:

The value of the resistance is given in the form:

R = AB × C ± D%

Where,

‘A’ and ‘B’ indicate the first two significant resistance figures (Ohms).

‘C’ indicates the decimal multiplies.

‘D’ indicates the tolerance in percentage.

The table for the resistor color code is given below:

Calculation

The resistor color code uses the following values for the first three bands:

Yellow → 4 (1st significant digit)

Purple (Violet) → 7 (2nd significant digit) 

Red → 10² = 100 (Multiplier)

Resistance = (First Digit × 10 + Second Digit) × Multiplier

Resistance = (4×10+7)×100 = 4700 Ω 

Temperature coefficient of resistance

The value of the resistance at temperature 't2' is given by:

\(R_t=R_o[1+α (t_2-t_1)]\)

where, Rt = Final resistance at temperature 't2'

Ro = Initial resistance at temperature 't1'

α = Temperature coefficient of resistance

Calculation

Given, R₄₀ ​= 4Ω at 40^oC

R80 ​= 6Ω at 80oC

Using the resistance-temperature formula:

\(4=R_{0}[1+α (40-0)]\) .....(i)

\(6=R_{0}[1+α (80-0)]\) .....(ii)

Multiplying equation (i) with 4 and equation (ii) with 2, we get:

\(16=R_{0}[4+160\alpha]\) .......(iii)

\(12=R_{0}[2+160\alpha]\) ........(iv)

Subtracting (iv) with (iii), we get:

\(2R_o=4\)

\(R_o=2\Omega \)

Concept

The resistance of a conductor is given by:

\(R={\rho l\over A}={4\rho l\over \pi d^2}\)

where, ρ = Resistivity

R = Resistance

l = Length of conductor

A = Cross-sectional area of the conductor

Observations:

• The resistance is directly proportional to the length of the conductor.
• The resistance is inversely proportional to the area of the cross-section of the conductor.
• The resistance depends upon the nature of the material.

Calculation

Given, d₂ = 0.5d₁

l₂ = 2l₁

\({R_2\over R_1}={l_2\over l_1}\times ({d_1\over d_2})^2\)

\({R_2\over R}={2l_1\over l_1}\times ({d_1\over 0.5d_1})^2\)

R₂ = 8R Ω

Explanation:

Effect of Removing a Resistor in a Parallel Circuit

Definition: In a parallel circuit, multiple resistors are connected in such a way that each resistor is connected to the voltage source independently. This means the voltage across each resistor is the same, but the total current flowing through the circuit is the sum of the currents through each resistor. The total resistance in a parallel circuit is found using the reciprocal formula:

Formula:

For resistors R₁, R₂, R_3, ..., R_n in parallel, the total resistance (R_total) is given by:

1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/R_n

Correct Option Analysis:

The correct option is:

Option 3: Increase

This option correctly describes the effect of removing one of the resistors (R1) from a parallel circuit. When a resistor is removed, the number of parallel paths decreases, which results in an increase in the total resistance of the circuit

The circuit after removing the voltage source

The total resistance of the new circuit will be the equivalent resistance of the network.

R_eq = R_t = 3 + 2 + 2 = 7 Ω 

The equivalent resistance of the network is 7 Ω.

 Mistake PointsWhile finding the equivalent resistance of the network, don't consider the internal resistance of the voltage source. Please read the question carefully it is mentioned in the question as well.

Concept:

The resistance of most of the metals increases in a linear way with temperature as shown and can be represented by the equation:

RT = R₀(1 + αT)

RT = Resistance at a temperature T

R₀ = Resistance at 0°C

α = Temperature coefficient of the resistance

Calculation:

Given R_40° = 4Ω and R_80° = 6Ω. We can write:

4 = R₀(1 + 40α)   ---(1)

6 = R₀(1 + 80α)   ---(2)

Multiplying Equation (1) with 4 and Equation (2) with 2, we get:

16 = R₀(4 + 160α)   ---(3)

12 = R₀(2 + 160α)   ---(4)

Subtracting (4) with (3), we get:

16 - 12 = 4R₀ - 2R₀

2R₀ = 4

R₀ = 2 Ω 

CONCEPT:

The resistance (R) of a conducting wire depends upon the length of wire, cross-sectional of the wire, and its resistivity.

Resistance is directly proportional to the length of the conductor and Inversely proportional to the cross-sectional area.

\(R = ρ \frac{l}{A}\)

Where l is the length of the conductor, a is the cross-sectional area of the conductor, ρ is resistivity. 

Resistivity: The resistivity is the resistor depends upon the nature of the conductor and temperature. ∴ Material 

Resistivity is the same for given conducting material at a given temperature and does not depend upon dimensions of wire.

If a wire is stretched, its volume remains the same. 

Concept:

The below equation shows the method to find the resistance:

Resistor Colour Coding uses colored bands to easily identify a resistor resistive value and its percentage tolerance. The resistor color code markings are always read one band at a time starting from the left to the right, with the larger width tolerance band oriented to the right side indicating its tolerance. 

The value of the resistance is given in the form:

R = AB × C ± D%

Where,

‘A’ and ‘B’ indicates the first two significant figures of resistance (Ohms).

‘C’ indicates the decimal multiplies.

‘D’ indicates the tolerance in percentage.

The table for the resistor colour code is given below:

Calculation:

From the above resistance colour codes,

Y → 4

V → 7

‘Y’ and ‘V’ defines ‘A’ and ‘B’

Red, R → 10²

‘Y’ defines C

Now, the resistance is:

R = 47 × 10² Ω 

R = 4700 ohms

Shortcut Trick

BB ROY Great Britain Very Good Watch Gold and Silver

B - Black (0)

B - Brown (1)

R - Red (2)

O - Orange (3)

Y - Yellow (4)

G - Green (5)

B - Blue (6)

V - Violet (7)

G - Grey (8)

W - Whilte (9)

Tolerance - Gold(5%) and Silver(10%)

Concept:

• Conductance is the measure of how easily electricity flows along a certain path through an electrical element. Represented with letter G.
• Conductance is the reciprocal of the Resistance and is measured in Siemens or mhos.

Equivalent Conductance in Series connection is represented as,

\({G_{eq}} = \frac{{{G_1}{G_2}}}{{{G_!} + {G_2}}}\)

Equivalent Conductance in parallel connection is represented as,

G_eq = G₁ + G₂

Calculation:

In the given figure,

6 S, 12 S are in series apply series combination formula

4 S, 8 S are in parallel apply parallel combination formula

Circuit reduces to,

Then 2 S and 4 S are in parallel add them,

Then, \({G_{eq}} = \frac{{12 \times 6}}{{12 + 6}}\; = 4\;S\)

CONCEPT: 

Resistance:

• The measurement of the opposition of the flow of electric current through a conductor is called resistance of that conductor. It is denoted by R.

There are mainly two ways of the combination of resistances:

1. Resistances in series:

• When two or more resistances are connected one after another such that the same current flows through them are called resistances in series.
• The net resistance/equivalent resistance (R) of resistances in series is given by:
• Equivalent resistance, R = R1 + R2

2. Resistances in parallel:

• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:

\(⇒\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\)

Calculation:

Reciprocal of resistance is conductance, we convert the conductances into resistances and the circuit will look like this

Given: R₁ = 1 Ω , R₂ = 1 / 3 Ω , R₃ = 1 / 2 Ω

Taking eqivalent resistance of R1 and R₂ 

\(R_{12} = \dfrac{1\times \frac13}{1+\frac13}=\frac14\; \Omega \)

Using current division:

I = \(I_{\frac12\Omega }=\dfrac{\frac14}{\frac14 + \frac12}\times 6 = 2\; A\)

Concept:

Conductance in Series:

\(\frac{1}{G_s}=\frac{1}{G_1}+\frac{1}{G_2}+\frac{1}{G_3}+......\frac{1}{G_n}\)

Conductance in Parallel:

G_P = G₁ + G₂ + G₃ + ..... G_n

Application:

Given circuit:

Here, 8 S is connected parallel with 4 S and, 6 S is connected series with 12 S

Hence, an equivalent circuit can be drawn as,

Now, 2 S is connected parallel with 4 S, equivalent circuit can be drawn as,

Hence,

\(\frac{1}{G_{eq}}=\frac{1}{12}+\frac{1}{6}=4\ S\)

CONCEPT:

Resistance: The obstruction offered to the flow of current is known as the resistance. It is denoted by R.

When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal then it is called resistances in parallel​

The net resistance/equivalent resistance(R) of resistances in parallel is given by:

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

EXPLANATION:

Given that,

R1 = low resistance 

R₂ = high resistance

Suppose R₁ = 3Ω, R₂ = 6Ω 

When the resistor is connected in parallel, then the equivalent resistance is

\(\frac{1}{{{R_{para}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} \)

∴ Rpara = (6 × 3)/9 = 2Ω 

Hence, it is proved that the combined resistance is smaller than the lower resistance. 

Concept:

Resistance at a temperature t is given by

R_t = R₀(1 + α Δt)

Where R_t = resistance at final temperature

R₀ = resistance at initial temperature

α = temperature coefficient

Δt = change in temperature

Calculation:

Given R_t = 48 ohm

R₀ = 45 ohm

α = 0.004/° C

48 = 45 (1 + 0.004 × Δt)

⇒ Δt = 16.67°

Δt = T₂ - T₁

Concept:

Conductance (G) is the reciprocal of Resistance (R).

Unit of Conductance is siemens (S).

Calculation:

Given:

S₁ = 6 S, S₂ = 5 S, S₃ = 20 S 

So, \({R_1} = \frac{1}{{{S_1}}}=\frac{1}{6}\;{\rm{\Omega }}\)

\({R_2} = \frac{1}{{{S_2}}} = \frac{1}{5}\;{\rm{\Omega }}\)

\({R_3} = \frac{1}{{{S_3}}} = \frac{1}{{20}}\;{\rm{\Omega }}\)

\({R_{23}} = {R_2} + {R_3} = \frac{1}{5} + \frac{1}{{20}} = \frac{5}{{20}} = \frac{1}{4}\;{\rm{\Omega }}\)

\({R_{eq}} = \frac{{{R_1} \times {R_{23}}}}{{{R_1} + {R_{23}}}} = \frac{{\frac{1}{6} \times \frac{1}{4}}}{{\frac{1}{6} + \frac{1}{4}}} = \frac{{\frac{1}{{24}}}}{{\frac{{10}}{{24}}}} = \frac{1}{{24}} \times \frac{{24}}{{10}} = \frac{1}{{10}}{\rm{\;\Omega }}\)

\({S_{eq}} = \frac{1}{{{R_{eq}}}} = 10\;S\)

Shortcut Trick:-

In series

• Resistance are added simple.
• Conductance are inversely (1/S) added.

In parallel

• Resistance are inversely (1/R) added.
• Conductance are added simple.

So, conductance S­₂ and S₃ are in series

Added inversely S₂₃ = (5*20)/(5+20)= 100/25 = 4 S

Now, S₁ and S₂₃  are parallel added simple

S_eq = 6+4 =10 S