resistors by nuhman10



    Application, types and characteristics of resistors. Simple DC circuits

Applications of resistors

In general, a resistor is used to create a known voltage-to-current ratio in an electric circuit. If
the current in a circuit is known, then a resistor can be used to create a known potential
difference proportional to that current. Conversely, if the potential difference between two
points in a circuit is known, a resistor can be used to create a known current proportional to
that difference.

Current-limiting. By placing a resistor in series with another component, such as a light-
emitting diode, the current through that component is reduced to a known safe value.

An attenuator is a network of two or more resistors (a voltage divider) used to reduce the
voltage of a signal.

All resistors dissipate heat. This is the principle behind electric heaters.

A resistor has a maximum working voltage and current above which the resistance may
change (drastically, in some cases) or the resistor may be physically damaged (overheat or
burn up, for instance). Most resistors are rated with a maximum power which is determined
by the physical size. Common power ratings for carbon composition and metal-film resistors
are 1/8 watt, 1/4 watt, and 1/2 watt. Metal-film and carbon film resistors are more stable than
carbon resistors against temperature changes and age. Larger resistors are able to dissipate
more heat because of their larger surface area. Wire-wound and resistors embedded in sand
(ceramic) are used when a high power rating is required.


A fuse (fusable link) is a type of over-current protection device. It has as its critical
component a metal wire or strip that will melt when heated by a prescribed (design) current,
opening the circuit of which it is a part, thereby protecting the circuit from an over-current
condition. Fuses are often characterized as "fast-blow" or "slow-blow" | "time-delay",
according to the time they take to respond to an over-current condition.

Types of resistor

                                   Fixed resistors

                                     Some resistors are cylindrical, with the actual resistive
                                     material in the centre (composition resistors), or on the
                                     surface of the cylinder (film) resistors, and a conducting
                                     metal lead projecting along the axis of the cylinder at each
                                     end(axial lead). There are carbon film and metal film
                                     resistors. The on the left shows a row of common resistors.
                                     Power resistors come in larger packages designed to
                                     dissipate heat efficiently. At high power levels, resistors
tend to be wire wound types. Resistors used in computers and other devices are typically
much smaller, often in surface-mount packages without wire leads. Resistors are built into
integrated circuits as part of the fabrication process, using the semiconductor as the resistor.

Variable resistors

The variable resistor is a resistor whose value can be adjusted by turning a shaft or sliding a
control. These are also called potentiometers or rheostats and allow the resistance of the
device to be altered by hand. Variable resistors can be inexpensive single-turn types or multi-
turn types with a helical element. Some variable resistors can be fitted with a mechanical
display to count the turns.

Other types of resistors

A metal oxide varistor (MOV) is a special type of resistor that changes its resistance with rise
in voltage: a very high resistance at low voltage (below the trigger voltage) and very low
resistance at high voltage (above the trigger voltage). It acts as a switch. It is usually used for
short circuit protection in power strips or lightning bolt "arrestors" on street power poles, or as
a "snubber" in inductive circuits.

A thermistor is a temperature-dependent resistor. There are two kinds, classified according to
the sign of their temperature coefficients:

A Positive Temperature Coefficient (PTC) resistor is a resistor with a positive temperature
coefficient. When the temperature rises the resistance of the PTC increases. PTCs are often
found in televisions in series with the demagnetizing coil where they are used to provide a
short-duration current burst through the coil when the TV is turned on. One specialized
version of a PTC is the polyswitch which acts as a self-repairing fuse.

A Negative Temperature Coefficient (NTC) resistor is also a temperature-dependent resistor,
but with a negative temperature coefficient. When the temperature rises the resistance of the
NTC drops. NTCs are often used in simple temperature detectors and measuring instruments.

A sensistor is a semicondutor-based resistor with a negative temperature coefficient, useful in
compensating for temperature-induced effects in electronic circuits.

Light-sensitive resistors change their resistance when exposed to light. They are used to
measure light intensity.

                                      Identifying resistors

Most axial resistors use a pattern of coloured stripes to indicate resistance. SMT ones follow a
numerical pattern.

4-band axial resistors

4 band identification is the most commonly used colour coding scheme
on all resistors. It consists of four coloured bands that are painted
around the body of the resistor. The first two numbers are the first two
significant digits of the resistance value, the third is a multiplier, and the fourth is the
tolerance of the value. Each colour corresponds to a certain number, shown in the chart
below. The tolerance for a 4-band resistor will be 2%, 5%, or 10%.

5-band resistors

More precise value resistors are available with 0.05%, 0.1%, 0.25% , 0.5% and 1% tolerance.
They are denoted with 5 strips, the fifth one is the temperature coefficient.

The Standard EIA Colour Code Table per EIA-RS-279 is as follows:

Colour 1st band 2nd band 3rd band (multiplier) 4th band (tolerance) Temp. Coefficient
Black 0         0        ×100
Brown 1         1        ×101                  ±1% (F)              100 ppm
Red    2        2        ×10                   ±2% (G)              50 ppm
Orange 3        3        ×10                                        15 ppm
Yellow 4        4        ×10                                        25 ppm
Green 5         5        ×10                   ±0.5% (D)
Blue 6          6        ×10                   ±0.25% (C)
Violet 7        7        ×10                   ±0.1% (B)
Gray 8          8        ×10                   ±0.05% (A)
White 9         9        ×10
Gold                     ×0.1                  ±5% (J)
Silver                   ×0.01                 ±10% (K)
None                                           ±20% (M)

SMT resistors

Surface-mount resistors are printed with numerical values in a code related to that used on
axial resistors. Standard-tolerance SMT resistors are marked with a three-digit code, in which
the first two digits are the first two significant digits of the value and the third digit is the
power of ten. For example, "472" represents "47" (the first two digits) multiplied by ten to the
power "2" (the third digit), i.e.                                               . Precision SMT
resistors are marked with a four-digit code in which the first three digits are the first three
significant digits of the value and the fourth digit is the power of ten.

                                      Simple DC circuits

Current limiting and regulation

E=24 V R = 50 .
We want to limit the current to I  Imax = 0.1 A and
change it with the variable resistor Rv. The maximum
value of the variable resistor is (a potentiometer) is 50 .
Calculate the resistance of the current-limiting resistor RL.
What is the range the current can be set?

(RL=190 ., Imin =24/290=0.083 A : 0.1 A  I  0.083 A )

Matching the load to the internal resistance of the source to get maximum power

Assume we have a voltage source of emf  and internal resistance Ri . What should be the
                             resistance of the load to get maximum power?

                                                                          2R
                                   P  I R, I 
                                                   Ri  R             ( Ri  R) 2

                                  P has its maximum value when dP/dR=0

                                                                                    R  Ri
                                 dP    ( R  R)  2R( Ri  R)
                                    2 i                     0        2
                                 dR           ( Ri  R) 4         Pmax 
                                                                         4 Ri

We get maximum power on a load of resistance equal to the internal resistance of the source.

Single-loop circuit. How much current flows in the circuit and in what direction and what is
the potential of each points with respect the point O ?
We connect the “common” input of a digital voltmeter to E, and the “V” input to B. What
voltage is shown?

                                                             1 = 15 V
                                                             2 = 13 V
                                                             R1 = 100 
                                                             R2 = 200 
                                                             R3 = 400 

First we assign a hypothetical current i to the loop shown by an arrow and go round the circuit
in the direction of the arrow. According to Ohm’s law, the drop of the potential across a
resistor R is RI. A connecting wire has zero resistance, so U(A) =U(O) = 0.

The potential drop across R1 is 100i, so U(B)=-100i.
We go from point B to C, these points are at the terminals of the first battery, and the potential
rises by 15 V. U(C)=-100i+15.
The potential drops across R2 by 200i, so U(D)=-100i+15-200i.
The potential drops again across R3. UE = -100i+15-200i-400i.
The potential rises across the second battery by 13 V. U(O) =-100i+15-200i-400i.+13. We
arrived back to point O, to zero potential, so

 100i  15  200i  400i  13  0  700i  28  i  0.04 A
I=0.4 A current flows in the direction noted by the arrow. Knowing the current, we get the
potentials: U(A)=0, U(B)=--4 V, U(C)=-4+15=11 V, U(D)=11-8=3 V, U(E)=3-16=-13 V,
U(O)-13+13 = 0 V

We connect the “common” input of a digital voltmeter to E, and the “V” input to B. What
voltage is shown?
The voltmeter reads the potential difference between its “V” input with respect to the “com”
Input. So the voltage shown is

V=UB-UE= -4 – (-13) = 9 V.

The loop method to solve circuits containing two or more loops

Determine the currents flowing across each resistor in the circuit shown in the picture. What
is the potential of point A with respect to O?

                                                     1 = 50 V
                                                     2 = 400 V
                                                     3 = 290 V
                                                     R1 = 500 
                                                     R2 = 200 
                                                     R3 = 100 
                                                     R4 = 300 

First we assign a current direction to the loops in the circuit. In a branch, which belongs two
loops, the difference of the loop current will flow. In this way, Kirchhoff’s first law
automatically fulfils. At A, for example, i1 current enters from the left, i2 flows out to the
right, and (i1-i2) flows out downward. The net current is i1-(i1-i2)-i2 = 0
Starting from a point of a loop we add up all the potential differences along the loop resulting
in zero when we reach back.

Along the left loop, starting from “O”, the potential drops on R1 by R1i1, then drops again by
3 across the battery, then drops again by R3(i1-i2) across R3.

 2  R1i1  ( 3 )  R2 (i1  i2 )  0
We get the equation for the right loop on the same way, starting form O, following the arrow
and adding up the changes of the potential:

 R2 (i2  i1 )   3  1  R3i2  R4i2  0
Arranging the equations, we get

                             ( R1  R2 )i1  R2i2   2   3
                            R2 i1  ( R2  R3  R4 )i2   1   3
Plugging in the data:

-700 i1+200 i2 = -110
200i1-600 i2 –240

The solution is i1=0.3 A, i2 = 0,5 A.

through R1 : 0.3 A, from left to right, in the direction shown by the arrow
through R2 : 0.3 A, 0.3 A downward and 0.5 A upward, the net current is 0,2 A upward.,
through R3 and R4: 0.5 A in the direction of the arrow.
The change of the potential from O to A is –R20.2 A +290 = 250 V. VAO = 250 V.

The voltage divider

                                  The emf of the source is  = 24 V. How do we ensure
                                  VAB=10 V?
                                                                    
                                  VAB  IR2              R2 
                                                R1  R2          1  R1 / R2

                                  The ratio of the resistors has to be R1/R2 = 1.4
                                  The voltage VAB can be regulated continuously between 0
                                  and  by a potentiometer.

The potentiometer is also used to measure voltage or detect
slight change of voltage in compensator circuits. A Pogendorf
Compensator is shown in the picture below.

Poggendorf Compensator
                                  A very sensitive galvanometer
                                  G detects if there is current
                                  flowing in the bottom circuit. The current can be zero if the
                                  voltage VS0 is he same as the unknown emf, x. If the
                                  current IG = 0 the total current Ip flows through the helical
                                  potentiometer RH, and VSO=IpR. Ip depends only on the
                                  resistance of the potentiometer and both on the emf and
                                  internal resistance of the source.
                                  A helical potentiometer is equipped with a 1000- division
                                  scale from where the position of the sliding contact can be
                                  read, and R=n/1000 RH . Let be the sliding contact at the
                                  division nx. We replace the voltage source with a standard

cell of accurately know emf, 0, and set the sliding contact for zero current again. Let it be n0.
x       RH I p
     1000              n
                  x  x 0
      n                n0
 0  0 RH I p
The Wheatstone Bridge is used to measure resistance by comparing an unknown resistance
with a standard one. It is also used to detect slight variation of resistance.

If the galvanometer reads zero the same current I1 flows
through both resistors R1 and R2 and again the same
current I2 flows through R3 and R4. Moreover, the
voltage across the galvanometer is zero, so A and B are at
the same potential. This means for the voltages, VAO and
VBO and VCA and VCB that VAO = VBO and VBO = VCB.
But these voltages are proportional to the current across
the resistors, VAO = I1R1 and VBO = I2R3 I1R1=I2R3. A
similar equation holds for the other pair of resistors R2
and R4: I1R2=I2R4. Dividing the equations by each other
we get the condition of bridge balance:

 R1 R3
 R2 R4
If one pair of resistors (R3,R4) are fixed, and their ratio is accurately known, and one of the
other two resistors (say R2) is a variable resistor Rv (a rheostat) we can determine the unlknow
resistance R1 from the value Rv set for balance: R1  Rv        .

Application of Thevenin’s Theorem for the Wheatstone Bridge

Use Thevenin’s theorem to calculate the current reading of the galvanometer in terms of Rx
near balance in the Wheatstone-Bridge in the figure. The galvanometer has got the internal
resistance RG and the battery can be considered an ideal voltage source with zero internal

First we determine the Thevenin equivalent emf and
internal resistance of the two-pole AB, with the
galvanometer removed.

Assuming the potential zero at O,

         R1                Rx
UA              , UB              , so the open-circuit voltage is
       R1  R2            Rx  Rv
                     R1         Rx 
                     R  R R  R   T
VAB  U A  U B                  
                     1    2   x    v
                                                             R1 Rx
When the Bridge is balanced, T = 0, and this holds when       
                                                             R2 Rv
The internal resistance of the Thevenin-equivalent voltage source is obtained as the resultant
resistance RAB when the battery is replaced with a short. We have the resistors Rx and Rv
connected in parallel then, and so are R1 and R2. The parallel resultants are connected in
series. The internal resistance is then

        R1R2          Rx Rv
RT                         .
       R1  R2       Rx  Rv
Connecting the galvanometer between A and B, the current flowing through it is
IG              .
       RT  RG

Measuring temperature with a resistance thermometer using a Wheatstone Bridge

Assume that Rx is a platinum resistance thermometer of resistance Ro =100  at T0 = 0C
      and temperature coefficient = 0.00386/C . The resistance changes with the
      temperature as R(T)=R0(!+T). The bridge is balanced with the variable resistor Rv
      at 0 C. When the resistance thermometer warms up, its resistance changes by r. A
      high-resistance voltmeter is connected between points A and B. Derive a formula to
      get the temperature from the voltage reading. Use linear approximation. Assume that
      R1=R2 and =12 V.

                                    R1         Rx 
From the previous problem, VAB   
                                    R  R R  R .
                                                   
                                    1    2   x    v
                                        R1 R0
It is zero at balance, when Rx = R0          Rv  R0
                                        R2 Rv

           r ( R0  Rv )  r  R0          r  Rv
VAB                                
                 ( R0  Rv ) 2            ( R0  Rv ) 2
                            r         r               V
Near balance: V AB           , but     T so T  4 AB .
                           4R0         R0               

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