In a resistor, the first three bands from left to right have colours yellow, violet and red. What is the value of the resistor in ohms.
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:
Colour code 
Values (AB) 
Multiplier (C) 
Tolerance (D) 
Black 
0 
100 

Brown 
1 
101 

Red 
2 
102 

Orange 
3 
103 

Yellow 
4 
104 

Green 
5 
105 

Blue 
6 
106 

Violet 
7 
107 

Grey 
8 
108 

White 
9 
109 

Gold 
 
 
± 5 % 
Silver 
 
 
± 10 % 
No color 
 
 
± 20 % 
Calculation:
From the above resistance colour codes,
Y → 4
V → 7
‘Y’ and ‘V’ defines ‘A’ and ‘B’
Red, R → 10^{2}
‘Y’ defines C
Now, the resistance is:
R = 47 × 10^{2} 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:
The resistance offered by a wire of unit length and unit area of crosssection is called resistivity or specific resistance (r).
Mathematically, this is defined as:
\(ρ=R.\frac{A}{l}\)
R is the resistance of the wire
A is the area of crosssection of the wire
l is the length of the wire.
Calculation:
With R = 0.171 Ω, l = 1 m and A = 0.1 mm^{2} = 0.1 × 10^{6} m^{2}
\(ρ=0.171× \frac{0.1× 10^{6}}{1}\)
ρ = 1.71 × 10^{8} Ωm
Calculate G_{eq} in the following circuit:
Concept:
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} = G_{1} + G_{2}
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:
The resistance of a conductor is given by:
\(R = \rho\frac{{ ~l}}{A}\)
ρ = resistivity
R = Resistance
l = length of wire
A = crosssectional 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 \(\frac{r_1}{r_2}=\frac{1}{2}\)
\(\frac{R_1}{R_2}=\frac{1}{4}\times (\frac{2}{1})^2\)
\(\frac{R_1}{R_2}=\frac{1}{1}\)
The approximate equivalent resistance between terminals A and B for the following infinite ladder network comprising of 1 Ω & 2 Ω resistors is:
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.
Resistance (R):
Resistance is a property of the conductor which opposes the current flow through it.
Resistance is measured in ohms and denoted by Ω.
Formula:
The resistance of a conductor measured as
\(R=\frac{ρ l}{A}\) Ω
ρ = Specific resistivity of the conductor
l = Length of the conductor
A = Crosssectional area of the conductor
\(A=\frac{\pi d^2}{4}\)
R ∝ (1 / A)
The thickness reminds the diameter of the conductor.
Calculation:
Let the resistance of a thick conductor is
R_{1} = 10 Ω
Another resistance is R_{2}
The ratio of the areas is
A_{1} : A_{2} = 3 : 1
R ∝ (1/ A)
\(\frac{R_2}{R_1}=\frac{A_1}{A_2}\)
⇒ R_{2 }= 3 x 10
∴ R_{2} = 30 Ω
The resistances are connected in series.
So, the equivalent resistance is
R_{eq} = R_{1} + R_{2}
⇒ R_{eq} = 10 + 30
∴ R_{eq} = 40 Ω
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 = R_{0}(1 + αT)
RT = Resistance at a temperature T
R_{0} = Resistance at 0°C
α = Temperature coefficient of the resistance
Calculation:
Given R_{40°} = 4Ω and R_{80°} = 6Ω. We can write:
4 = R_{0}(1 + 40α) (1)
6 = R_{0}(1 + 80α) (2)
Multiplying Equation (1) with 4 and Equation (2) with 2, we get:
16 = R_{0}(4 + 160α) (3)
12 = R_{0}(2 + 160α) (4)
Subtracting (4) with (3), we get:
16  12 = 4R_{0}  2R_{0}
2R_{0} = 4
R_{0} = 2 Ω
Two materials A and B have resistance temperature coefficients 0.004 and 0.0004 per °C respectively at a given temperature. In what proportion should A and B be joined in series to produce a circuit having a temperature coefficient of 0.001.
Concept:
Effect of Temperature on Resistance:
Temperature Coefficient of Resistance:
Let a metallic conductor having resistance R_{0} at 0°C be heated at t°C and resistance become R_{t}.
Increased in temperature = ΔR = R_{t} – R_{0}
An increase in Resistance (ΔR) depends on:
∴ (R_{t} – R_{0}) α (R × t)
or, (R_{t} – R_{0}) = α R_{0} t (1)
Where α is constant and known as the temperature coefficient of resistance.
From equation (1)
\(α = \frac{{{R_t}  {R_o}}}{{{R_o} × t}}\)
∴ R_{t} = R_{0} (1 + α t)
Calculation:
Let the resistance of material A (R_{A}) = 1 Ω .... (1)
And Resistance of material B (RB) = x Ω .... (2)
∴ Resistance of series combination (R_{c}) = (1 + x) Ω
Consider temperature rise = t° C
Hence, the resistance of series combination at t° C will be, (R_{ct}) is given as,
R_{ct} = (1 + x)(1 + 0.001t) .... (3)
Similarly,
Resisistance of material A at t° C will be,
R_{At} = 1(1 + 0.004t) .... (4)
And, Resisistance of material B at t° C will be,
R_{Bt} = x(1 + 0.0004t).... (5)
Since, both the resistance is inseries,
∴ R_{At} + R_{Bt} = R_{ct}
or, 1(1 + 0.004t) + x(1 + 0.0004t) = (1 + x)(1 + 0.001t)
or, 0.004t + 0.0004xt = (1 + x) × (0.001t)
Divided by t on both side,
0.004 + 0.0004x = (1 + x) × (0.001)
or, 0.004 + 0.0004x = 0.001 + 0.001x
or, 0.003 = 0.0006x
∴ x = 5
From equation (2),
Resistance of material B (RB) = x Ω = 5 Ω
∴ R_{A} : R_{B} = 1 : 5
Alternate Method
Consider the resistance of material A is R_{A }and the temperature coefficient is α_{A }and the resistance of material B is R_{B }and the temperature coefficient is α_{B}.
∴ \({α _A} = 0.004\;per^\circ C\;and\;{α _B} = 0.0004\;per^\circ C\)
Given the overall value of temperature coefficient when connected in series is α = 0.001 per ° C
By using concepts of Allegation and Mixture
\(\alpha _A^{new} = {\alpha _A}  \alpha = 0.004  0.001 = 0.003\;per^\circ C\)
And,
\(\alpha _A^{new} = {\alpha _A}  \alpha = 0.004  0.001 = 0.003\;per^\circ C\)
Therefore, the ratio of the resistance of twowire will be:
\(\frac{{\alpha _A^{new}}}{{\alpha _B^{new}}} = \frac{{0.003}}{{0.0006}} = 5\)
A resistance of 10 Ω is connected in series with two resistance, each of 15 Ω arranged in parallel. What resistance must be shunted across this parallel combination so that the total current taken shall be 1.5 A with 20 V applied voltage?
Drawing the circuit diagram as per the question,
Given V = 20 V and I = 1.5 A
So, equivalent resistance = (V / I) .......(1)
Also, from circuit equivalent resistance = 10 + (7.5  R) .........(2)
Equating equation (1) and (2)
\(\large{\frac{20}{1.5}=10+\frac{7.5R}{7.5+R}}\)
\(\large{\frac{40}{3}=10+\frac{7.5R}{7.5+R}}\)
\(\large{\frac{10}{3}=\frac{7.5R}{7.5+R}}\)
12.5 R = 75
R = 6 Ω
Concept:
Mathematically, the resistance is given by the formula:
\(R=ρ\frac{l}{A}\)
ρ = Resistivity of the material
l = length of the material
A = Area of the crosssection
Since A = πr^{2}, the above expression becomes:
\(R=\frac{\rho l}{\pi r^2} = \frac{4\rho l}{\pi d^2}\)
d = diameter of the wire
Application:
Initially, for a length 'l_{1}' and diameter of 'd_{1}', let the resistance be R_{1}, i.e.
\(R_1= \frac{4\rho l_1}{\pi d_1^2}\) (1)
After the wire is stretched by 5 times and the diameter is reduced to half, we have:
l_{2} = 5l_{1} and
\(d_2=\frac{d_1}{2}\)
The resistance will now be:
\(R_2= \frac{4\rho l_2}{\pi d_2^2}=\frac{4\rho (5l_1)}{\pi (d_1/2)^2}\)
\(R_2= \frac{80\rho l_1}{\pi d_1^2}\) (2)
Dividing Equation (2) and (1), we get:
\(\frac{R_2}{R_1}=\frac{80}{4}\)
R_{2} = 20 times of R_{1}
CONCEPT:
Resistance:
There are mainly two ways of the combination of resistances:
1. Resistances in series:
2. Resistances in parallel:
\(⇒\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\)
Calculation:
Reciprocal of resistance is conductance, we convert the conductances into resistances and the circuit will look like this
Given: R_{1} = 1 Ω , R_{2} = 1 / 3 Ω , R_{3} = 1 / 2 Ω
Taking eqivalent resistance of R1 and R_{2}
\(R_{12} = \dfrac{1\times \frac13}{1+\frac13}=\frac14\; \Omega \)
Using current division:
I = \(I_{\frac12\Omega }=\dfrac{\frac14}{\frac14 + \frac12}\times 6 = 2\; A\)
Concept:
Mathematically, Resistance is defined as:
\(R=\rho \frac{L}{A}\)
R is the resistance of the wire
A is the area of crosssection 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 I_{1} = 80 % more than I_{2} = 1.8 I_{2}
Length L_{2} = 45 % longer than L_{1} = 1.45 L_{1}
ρ_{1} : ρ_{2} = 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 \(\frac{R_2}{R_1}=\frac{\rho_2 \frac{L_2}{A_2}}{\rho_1 \frac{L_1}{A_1}}= \frac{\rho_2L_2A_1}{\rho_1L_1A_2}\)
\(\Rightarrow 1.8 =\frac{\rho_2L_2A_1}{\rho_1L_1A_2}=\frac{20\;\times\;1.45L_1A_1}{40\;\times\;L_1A_2}\)
\(\Rightarrow \frac{A_2}{A_1}=0.402\)
Concept:
Resistance: The property by which an electric conductor opposes the flow of current through it is called the resistance of the conductor.
It is denoted by R.
The SI unit of resistance is Ohm (Ω).
\(R = \frac{{ρ \;l}}{A}\)
Where ρ is the resistivity of a conductor, l is the length of the conductor and A is the crosssectional area.
Calculation:
When the cable is 700 m long, the resistance of the connected loop is found to be 100 Ω.
Let,
l_{1} = 700 m
R_{1} = 100 Ω
For 8 km long cable,
l_{2} = 8 km = 8000 m
From the above concept,
\(R∝ l\)
∴ 100 Ω ∝ 700 m .... (1)
R_{2} ∝ 8000 m .... (2)
From equation (1) and (2),
\(\frac{R_2}{100}=\frac{8000}{700}\)
R_{2} = 1,143 Ω
Concept:
Resistance,
\(R=\frac{ρ\ l}{A}\) .......(1)
Where ρ = resistivity of the conductor (Ω m)
l = length of conductor (m)
A = crosssection area of conductor (m^{2})
A = π r^{2} .........(2)
Where r is the radius of the crosssection of the conductor.
Explanation:
From equations (1) and (2),
\(R\propto\frac{1}{r^2}\) as length 'l' and resistivity 'ρ' is constant.
\(\large{\frac{R_1}{R_2}=\frac{r_2^2}{r_1^2}}\)
Diameter is doubled means radius is doubled.
Given, R_{1} = 400 Ω , r_{2} = 2 r_{1}
Where,
R_{1}, r_{1} is the old resistance and old radius of the wire.
R_{2}, r_{2} is the new resistance and new radius of the wire.
\(\large{\frac{400}{R_2}=\frac{(2r_1)^2}{(r_1)^2}=4}\)
R_{2} = 100 Ω
Therefore, new resistance R_{2} = 100 Ω
Find the value of G_{eq} in the following circuit.
10 S
Concept:
Conductance (G) is the reciprocal of Resistance (R).
Unit of Conductance is siemens (S).
Calculation:
Given:
S_{1} = 6 S, S_{2} = 5 S, S_{3} = 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
In parallel
So, conductance S_{2 }and S_{3 }are in series
Added inversely S_{23 }= (5*20)/(5+20)= 100/25 = 4 S
Now, S_{1 }and S_{23 } are parallel added simple
S_{eq }= 6+4 =10 S
Concept:
Colour Coding of Resistance: To know the value of resistance colour code is used:
Colour band A and B: Indicate the first two significant figures of resistance in ohms.
Band C: Indicates the decimal multiplier i.e. the number of zeros that follows the two significant figures.
Band D: Indicates the tolerance in percentage about the indicated value or in other words, it represents the percentage accuracy of the indicated value.
The tolerance in the case of gold is ±5% and in silver is ±10%. If only three bands are marked on carbon resistance, then it indicates a tolerance of 20%.
The table for the resistor colour code is given below:
Colour code 
Values (AB) 
Multiplier (C) 
Tolerance (D) 
Black 
0 
100 

Brown 
1 
101 
1 
Red 
2 
102 
2 
Orange 
3 
103 

Yellow 
4 
104 

Green 
5 
105 
0.5 
Blue 
6 
106 
0.25 
Violet 
7 
107 
0.1 
Grey 
8 
108 

White 
9 
109 

Gold 
 
 
\(\pm 5{\rm{\% }}\) 
Silver 
 
 
\(\pm 10{\rm{\% }}\) 
No colour 
 
 
\(\pm 20{\rm{\% }}\) 
Calculation:
Given: A, B, and C of Orange colour. From the table, we get:
R = (AB) × C
R = 33 × 10^{3} Ω
R = 33 kΩ
Standard resistances wire:
Rating of resistor:
Resistance is an element of a circuit or network which is inserted to lower the current in the circuit.
The unit of resistance is ohm and its symbol is Ω.
Usually, voltage V is constant 230 V (RMS) in household supply and wattage, P = V^{2}/R
So, if we fix the value of R, Wattage(Power) will be fixed.
Or we can say if we know anyone between resistance(R) and wattage (P), others can be known.
Thus, the rating is ohmic and wattage for a resistor.
Concept:
Mathematically, the resistance is given by the formula:
\(R=ρ\frac{l}{A}\)
ρ = Resistivity of the material
l = length of the material
A = Area of the crosssection
Since A = πr2, the above expression becomes:
\(R=\frac{\rho l}{\pi r^2} = \frac{4\rho l}{\pi d^2}\)
d = diameter of the wire
Application:
Initially, for a length 'l1' and diameter of 'd1', let the resistance be R1, i.e.
\(R_1= \frac{4\rho l_1}{\pi d_1^2}\) (1)
After the wire is stretched 3 times and the diameter is reduced to half, we have:
l2 = 3l1 and
\(d_2=\frac{d_1}{2}\)
The resistance will now be:
\(R_2= \frac{4\rho l_2}{\pi d_2^2}=\frac{4\rho (3l_1)}{\pi (d_1/2)^2}\)
\(R_2= \frac{48\rho l_1}{\pi d_1^2}\) (2)
Dividing Equation (2) and (1), we get:
\(\frac{R_2}{R_1}=\frac{48}{4}\)
R2 = 12 times of R1
The correct answer is 1 volt / 1 ampere.
Key Points
Quantity  SI Unit 
Frequency  Hertz 
Electric Potential  Volt 
Capacitance  Farad 
Magnetic Flux  Weber 
Power  Watt 
Magnetic Flux Density  Tesla 
Work  Joule 
Angle  Radians 
Inductance  Henry 
Electric Charge  Coulomb 
Additional Information
Quantity  SI Unit 
Length  Meter 
Amount of Substance  Mole 
Mass  Kilogram 
Luminous Intensity  Candela 
Time  Second 
Thermodynamic Temperature  Kelvin 
Electric Current  Ampere 
Electric Resistance/ Electric Impedance/ Electric Reactance  Ohm 
Radioactivity  Becquerel 
Electrical Conductance  Siemens 