If the temperature of a conductor remains constant, then the ratio of thepotential difference (or voltage) V to the current I remain constant for any conductor, which is known as resistance of the conductor.
\(\frac{V}{I}\) = constant = R
The resistance ‘R’ is measured in ohm (). Ohm is the resistance of that conductor
which carries one ampere of current when the potential difference between its ends
is one volt.
\(1\,ohm=\frac{1volt}{1amp}\)\(=\frac{{{10}^{8}}emu\,of\,potential}{{{10}^{-1}}emu\,of\,current}\)
=\({{10}^{9}}emu\,\) of resistance.
Ohms law is just an empirical relationship. It is just the relationship between
p.d and current
V = iR applicable for both ohmic and non ohmic substances.
Dimensional formula\(\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-2}} \right]\)
The conductors which obey Ohm’s law are called Ohmic conductors.
Ø All metals obey Ohm’s law.
Ø For Ohmic conductors V – i graph is a straight line passing through origin (metals).
Ex: Thermistor, Electronic Valve, Semi-conductor devices, gases, crystal rectifier etc.,
nonohmic Circuits:
Ø The circuits in which Ohm's law is not obeyed are called nonohmic circuits. The V-I graph is a curve, e.g. torch bulb, electrolyte,
semiconductors, thermonic valves etc. as shown by curves (a), (b), (c).
Ø the resistance of a conductor depends upon
1) shape (dimensions)
2) nature of material
3) impurities
4) Temperature
Ø the resistance of a conductor increases with impurities.
Ø The resistance of a semiconductor decreases with impurities.
Temperature dependence of resistance:
For conductors i.e metals resistance increases with rise in temperature
If \({{\text{R}}_{\text{0}}}\) = resistance of conductor at 0oC
If \({{\text{R}}_{\text{t}}}\) = resistance of conductor at toC.
And =\(\alpha ,\beta \) temperature co-efficient of resistance then
\({{R}_{T}}={{R}_{\circ }}\left( 1+\alpha t+\beta {{t}^{2}} \right)fort>{{300}^{\circ }}C\)
\({{R}_{T}}={{R}_{\circ }}\left( 1+\alpha t \right)fort<{{300}^{\circ }}C\)
\(\alpha =\frac{{{R}_{t}}-{{R}_{{}^\circ }}}{{{R}_{{}^\circ }}t}{}^\circ C\)
If \({{R}_{1}}\) and\({{R}_{2}}\) are the resistances at t1oC to t2oC respectively then \(\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1+\alpha {{t}_{1}}}{1+\alpha {{t}_{2}}}\)
\(\therefore \alpha =\frac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}{{t}_{2}}-{{R}_{2}}{{t}_{1}}}\)
The value of is different at different temperatures.
At a given temperature \(\alpha =\frac{1}{{{R}_{t}}}\left( \frac{dR}{dt} \right)\) at t0C
Note : In general \({{R}_{0}}={{R}_{20}}\)
Graph shows the variation of resistivity with temperature for conductors,
semiconductors and for alloys like managing and constantan. Since the
resistivity of managing and constantan remains constant with respect to
change in temperature, these materials are used for the bridge wires and
resistance coils.
Circuit diagram:
Example:
1.Electric field (E) and current density (J) have relation
Solution:
\(E\infty \frac{1}{j}\)
2.Ohm’s Law is true
1) For metallic conductors at low temperature
2) For metallic conductors at high temperature
3 For electrolytes when current passes through them
4) For diode when current flows.
Solution:
Ohm’s Law is true for metallic conductors at low temperature.
1.Using three wires of resistances 1 ohm, 2ohm and 3 ohm, then no.of different values of resistances that possible are?
Solution:
no of combinations x = 2n.
2.Current ‘i’ coming from the battery and ammeter reading are
Solution:
\(\frac{3}{8}A\frac{1}{8}A\) .
1.Determine the potential difference between the ends of a wire of resistance
5W if 720C passes through it per minute.
Solution:
First, we determine the current, I = Q/t, or I = 720 C/60s = 12A.
Then by using Ohm’s law, V = IR, or V = (12A) (5W) = 60V.
2.The effective resistance across the points A and I is
Solution: The equivalent circuit is shown
\(1\,\Omega \)
1.A cylindrical conductor of length l and inner radius R1 and outer radius R2 has specific resistance r. A cell of emf e is connected across the two lateral faces of the conductor.Find the current drown from the cell.
Solution:
To calculate current through the conductor we have to calculate effective resistance Between its two ends. So, we will Consider the differential element of the cylindrical shell having radius x and thickness dx as shown in the figure.
dR = \(\rho \,\frac{dx}{2\,\pi \,x\ell }\)
\(\therefore R=\int\limits_{{{R}_{1}}}^{R_{2}^{{}}}{\rho \frac{dx}{2\pi xl}}(\because R=\rho \frac{l}{a})\)
Þ R = \(\frac{\rho }{2\pi l}\ln \left( {}^{{{R}_{2}}}/{}_{{{R}_{1}}} \right)\)
I = \(\frac{\varepsilon }{R}\)Þ I = \(\frac{2\pi l\varepsilon }{\rho \ln \left( {{\frac{{{R}_{2}}}{{{R}_{1}}}}_{{}}} \right)}\)
2.In the figure, the value of resistance to be connected between C and D so that The resistance of the entire circuit between A and B does not change with the number of elementary sets used is
Solution:
Resitance between AB is independent of number of sets used. Let x be the between AB.
\(A=R\parallel \left[ 2R+\frac{Rr}{R+r} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\to \left( 1 \right)\)
\(x=R\parallel \left[ 2R+x \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\to \left( 2 \right)\) Solve for r
Resistance is the property of a substance by which it opposes the flow of current through it.
For a given body, the resistance is defined as the ratio of applied potential difference to the resulting current,
R = \(\frac{V}{I}\)
This property of the substance is due to frequent collisions of electrons with the atoms of the conductor. SI unit of resistance is named ohm which is equivalent to V/A. Its dimensions are
[R] = \(\left[ \frac{V}{I} \right]=\left[ \frac{W}{qI} \right]=\frac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ AT \right]\,\,\,\left[ A \right]}=\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-2}} \right]\)
Reciprocal of the resistance is called conductance (G),
G = \(\frac{1}{R}=\frac{I}{V}\).
with dimensions [M-1L-2T3A2]. SI unit of conductance is called siemens (S), which
is equivalent to ohm-1.
The electrical resistance of an electrical conductor is a measure of the difficulty to pass an electric current through that conductor. The inverse quantity is electrical conductance and is the ease with which an electric current passes. Electrical resistance shares some conceptual parallels with the notion of mechanical friction. The SI unit of electrical resistance is the ohm (Ω), while electrical conductance is measured in siemens (S).
An object of uniform cross section has a resistance proportional to its resistivity and length and inversely proportional to its cross-sectional area. All materials show some resistance, except for superconductors, which have a resistance of zero.
The resistance (R) of an object is defined as the ratio of voltage across it (V) to current through it (I), while the conductance (G) is the inverse:
\(R=\frac{V}{I},G=\frac{I}{V}=\frac{I}{R}\)
For a wide variety of materials and conditions, V and I are directly proportional to each other, and therefore R and G are constant(although they can depend on other factors like temperature or strain). This proportionality is called Ohm's law, and materials that satisfy it are called ohmic materials.In other cases, such as a transformer, diode or battery, V and I are not directly proportional. The ratio V/I is sometimes still useful, and is referred to as a "chordal resistance" or "static resistance",[1][2] since it corresponds to the inverse slope of a chord between the origin and an I–V curve. In other situations, the derivative \(\frac{dv}{dI}\) may be most useful; this is called the "differential resistance".
For a given body, its resistance is found to be
That is, R µ L and R µ \(\frac{1}{S}\)
or R µ \(\frac{L}{S}\) = r \(\frac{L}{S}\)
where r is a constant called resistivity?
Fractional Change in Resistance
The resistance of a wire depends on its length L and cross-sectional area S (or, as S = pr2, on radius r). If the wire is stretched by a small amount, its resistance R changes.
Suppose the length L increases by a small amount DL. Then, we can write
R = r \(\frac{L}{S}\) = r \(\left[ \frac{{{L}^{2}}}{V} \right]\) (as LS = volume, V = constant)
Hence fractional change in resistance with a small change in length is
\(\frac{\Delta R}{R}=2\frac{\Delta L}{L}\)
Case II:
Suppose the wire is stretched so that the cross-sectional area S reduces by small amount DS. Then we can write
R = r \(\frac{L}{S}=\rho \frac{V}{{{S}^{2}}}\)
Hence \(\frac{\Delta R}{R}=-2\frac{\Delta S}{S}\).
Case III:
Suppose the wire is stretched so that the radius reduces by Dr. Then, we can write
R = r \(\frac{L}{S}=\rho \frac{LS}{{{S}^{2}}}=\rho \frac{V}{{{\left( \pi {{r}^{2}} \right)}^{2}}}=\frac{\rho V}{{{\pi }^{2}}}\frac{1}{{{r}^{4}}}\)
\\(\therefore \frac{\vartriangle R}{R}=-4\frac{\vartriangle r}{r}\)
Temperature dependence of resistance:
For conductors i.e metals resistance increases with rise in temperature
If \({{\text{R}}_{\text{0}}}\) = resistance of conductor at 0oC
If \({{\text{R}}_{\text{t}}}\) = resistance of conductor at toC.
And =\(\alpha ,\beta \) temperature co-efficient of resistance then
\({{R}_{T}}={{R}_{\circ }}\left( 1+\alpha t+\beta {{t}^{2}} \right)fort>{{300}^{\circ }}C\)
\({{R}_{T}}={{R}_{\circ }}\left( 1+\alpha t \right)fort<{{300}^{\circ }}C\)
\(\alpha =\frac{{{R}_{t}}-{{R}_{{}^\circ }}}{{{R}_{{}^\circ }}t}{}^\circ C\)
If \({{R}_{1}}\) and \({{R}_{2}}\) are the resistances at t1oC to t2oC respectively then \(\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1+\alpha {{t}_{1}}}{1+\alpha {{t}_{2}}}\)
\(\therefore \alpha =\frac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}{{t}_{2}}-{{R}_{2}}{{t}_{1}}}\)
The value of is different at different temperatures.
At a given temperature \(\alpha =\frac{1}{{{R}_{t}}}\left( \frac{dR}{dt} \right)\) at t0C.
Note : In general \({{R}_{0}}={{R}_{20}}\)
Graph shows the variation of resistivity with temperature for conductors,semiconductors and for alloys like managing and constantan. Since the resistivity of managing and constantan remains constant with respect to change in temperature, these materials are used for the bridge wires and resistance coils.
The resistivity of managing and constantan is almost independent of temperature.
Variation of resistance of some materials:
Strain dependence :
Just as the resistance of a conductor depends upon temperature, the resistance of a conductor depends upon strain By placing a conductor under tension (a form of stress that leads to strain in the form of stretching of the conductor), the length of the section of conductor under tension increases and its cross-sectional area decreases. Both these effects contribute to increasing the resistance of the strained section of conductor. Under compression (strain in the opposite direction), the resistance of the strained section of conductor decreases.
Light illumination dependence:
Some resistors, particularly those Made from semiconductors,exhibit photoconductivity, meaning that their resistance changes when light is shining on them. Therefore, they are called photoresistors (or light dependent resistors). These are a common type of light ditector.
1.From the following the quantity which is analogous to temperature in electricity is
Solution:
The quantity which is analogous to temperature in electricity is potential.
2.The resistance of a conductor is
1) Inversely proportional to the length.
2) Directly proportional to the square of the radius.
3) Inversely proportional to the square of the radius.
4) Directly proportional to the square root of the length.
Solution:
The resistance of a conductor is Inversely proportional to the square of the radius.
1.A wire 50 cm long and 1mm2 in cross-section carries a current of 4A when connected to a 2V battery. The resistivity of the wire is
Solution:
\(V=\frac{i\rho l}{A}\Rightarrow \rho =\frac{VA}{il}\)
\(1\times {{10}^{-6}}\,\,\Omega m\)
2.A wire of resistance 20 is bent in the form of a square. The resistance between the ends of diagonal is
Solution:
\({{R}^{1}}=\frac{20}{4}=5\)
\({{R}_{1}}=10\Omega ,\,{{R}_{2}}=10\)
\({{R}_{P}}=\frac{10}{2}=5\)
1.A uniform wire of resistance \(1\ \Omega /m\) having resistance is bent in the form of circle as shown in fig. If the equivalent resistance between M and N is\(1.8\ \Omega \) ,then the length of the shorter section is
Solution:
Let the resistance of shorter part MN be x.
Then resistance of longer part is (20 – x)\(\Omega \)
\({{R}_{eq}}=\frac{(20-x)x}{20-x+x}=1.8\ \Omega \)
Solving we get x =\(2\ \Omega \)
So length of shorter part = 2m
2.Twelve resistances each of resistance R are connected in the circuit as shown in fig. Net resistance between points A and Cwould be
Solution:
When a cell is connected across A and C, no current flows in the arms BG and ED due to symmetry in the arrangement. Then equivalent circuit will be as shown in Fig.(a) and (b). The effective resistance between A and C is
\(\frac{1}{{{R}_{eff}}}=\frac{1}{2R}+\frac{1}{3R}+\frac{1}{2R}\)
\(=\frac{3+2+3}{6R}=\frac{8}{6R}=\frac{4}{3R}\) or \({{R}_{eff}}=\frac{3R}{4}\)
1.A wire has a resistance R. What will be its resistance,
(a) if it is doubled on itself,
(b) if it is stretched so that
(i) its length is doubled
(ii) its radius is halved ?
Solution:
For a wire,
R = r \(\frac{L}{S}=\rho \frac{L}{\pi {{r}^{2}}}\)
If its dimensions are changed without changing its mass, its volume remains constant,
V = V¢ or SL = S¢L¢
(a) When the wire is doubled on itself, its length is halved and its area of cross-section is doubled.
S ´ L = S¢ ´ (L/2) or S¢ = 2S
\R¢ = \(\rho \frac{L'}{S'}=\rho \frac{\left( L/2 \right)}{\left( 2S \right)}=\frac{1}{4}\rho \frac{L}{S}=\frac{1}{4}R\)
(b) (i) When the length of the wire is doubled by stretching it,
S ´ L = 2L ´ S¢ or S¢ = S/2
\ R¢ = r \(\frac{L'}{S'}=\rho \frac{2L}{S/2}=\rho \frac{4L}{S}=4\rho \frac{L}{S}=4R\)
(ii) When the wire is stretched so as to reduce its radius to half (r¢ = r/2), we have
S¢ = pr¢2 = p (r/2)2 = pr2/4 = S/4
From S ´ L = S¢ ´ L¢ = (S/4) ´ L¢
Þ L¢ = 4L
\R = r\(\frac{{{L}^{'}}}{{{S}^{'}}}=\rho \frac{4L}{S/4}=16\rho \frac{L}{S}=16R\).
2.A metallic sphere of radius a is surrounded by a concentric thin metallic sheet of radius b. The space between these two electrodes is filled with a homogeneous poorly conducting material of resistivity r.
(a) Find the resistance between the two electrodes.
(b) Also derive an expression for the current density as a function of radius r, if the potential difference between the electrodes is V.
Solution:
(a) Consider a thin spherical layer between radii r and r + dr. The lines of current at all points of this layer are perpendicular to it. Therefore, such a layer can be treated as a conductor of length dr and cross-sectional area 4pr2. Thus,
dR = r \(\frac{dr}{4\pi {{r}^{2}}}\)
\R = \(\int\limits_{a}^{b}{dR}=\int\limits_{a}^{b}{\frac{\rho \text{ }dr}{4\pi {{r}^{2}}}}\)
\(=\frac{\rho }{4\pi }\left[ \frac{-\text{1}}{r} \right]_{a}^{b}=\frac{\rho }{4\pi }\left( \frac{1}{a}-\frac{1}{b} \right)\)
(b) The current density is given by
\(J=\frac{I}{S}=\frac{V}{RS}\)
At a radius r, S = 4pr2.
\ \(J\text{ }(r)=\frac{V}{4\pi {{r}^{2}}}\frac{4\pi }{\rho }\left( \frac{a\text{ }b}{b-a} \right)\)\(=\frac{V}{\rho {{r}^{2}}}\left( \frac{a\text{ }b}{b-a} \right)\).
Resistance is the property of a substance by which it opposes the flow of current through it.
For a given body, the resistance is defined as the ratio of applied potential difference to the resulting current,
R = \(\frac{V}{I}\)
This property of the substance is due to frequent collisions of electrons with the atoms of the conductor. SI unit of resistance is named ohm which is equivalent to V/A. Its dimensions are
[R] = \(\left[ \frac{V}{I} \right]=\left[ \frac{W}{qI} \right]=\frac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ AT \right]\,\,\,\left[ A \right]}=\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-2}} \right]\)
Reciprocal of the resistance is called conductance (G),
G = \(\frac{1}{R}=\frac{I}{V}\).
with dimensions [M-1L-2T3A2]. SI unit of conductance is called siemens (S), which
is equivalent to ohm-1.
The electrical resistance of an electrical conductor is a measure of the difficulty to pass an electric current through that conductor. The inverse quantity is electrical conductance and is the ease with which an electric current passes. Electrical resistance shares some conceptual parallels with the notion of mechanical friction. The SI unit of electrical resistance is the ohm (Ω), while electrical conductance is measured in siemens (S).
An object of uniform cross section has a resistance proportional to its resistivity and length and inversely proportional to its cross-sectional area. All materials show some resistance, except for superconductors, which have a resistance of zero.
The resistance (R) of an object is defined as the ratio of voltage across it (V) to current through it (I), while the conductance (G) is the inverse:
\(R=\frac{V}{I},G=\frac{I}{V}=\frac{I}{R}\)
For a wide variety of materials and conditions, V and I are directly proportional to each other, and therefore R and G are constant(although they can depend on other factors like temperature or strain). This proportionality is called Ohm's law, and materials that satisfy it are called ohmic materials.In other cases, such as a transformer, diode or battery, V and I are not directly proportional. The ratio V/I is sometimes still useful, and is referred to as a "chordal resistance" or "static resistance",[1][2] since it corresponds to the inverse slope of a chord between the origin and an I–V curve. In other situations, the derivative \(\frac{dv}{dI}\) may be most useful; this is called the "differential resistance".
For a given body, its resistance is found to be
That is, R µ L and R µ \(\frac{1}{S}\)
or R µ \(\frac{L}{S}\) = r \(\frac{L}{S}\)
where r is a constant called resistivity?
Fractional Change in Resistance
The resistance of a wire depends on its length L and cross-sectional area S (or, as S = pr2, on radius r). If the wire is stretched by a small amount, its resistance R changes.
Suppose the length L increases by a small amount DL. Then, we can write
R = r \(\frac{L}{S}\) = r \(\left[ \frac{{{L}^{2}}}{V} \right]\) (as LS = volume, V = constant)
Hence fractional change in resistance with a small change in length is
\(\frac{\Delta R}{R}=2\frac{\Delta L}{L}\)
Case II:
Suppose the wire is stretched so that the cross-sectional area S reduces by small amount DS. Then we can write
R = r \(\frac{L}{S}=\rho \frac{V}{{{S}^{2}}}\)
Hence \(\frac{\Delta R}{R}=-2\frac{\Delta S}{S}\).
Case III:
Suppose the wire is stretched so that the radius reduces by Dr. Then, we can write
R = r \(\frac{L}{S}=\rho \frac{LS}{{{S}^{2}}}=\rho \frac{V}{{{\left( \pi {{r}^{2}} \right)}^{2}}}=\frac{\rho V}{{{\pi }^{2}}}\frac{1}{{{r}^{4}}}\)
\\(\therefore \frac{\vartriangle R}{R}=-4\frac{\vartriangle r}{r}\)
Temperature dependence of resistance:
For conductors i.e metals resistance increases with rise in temperature
If \({{\text{R}}_{\text{0}}}\) = resistance of conductor at 0oC
If \({{\text{R}}_{\text{t}}}\) = resistance of conductor at toC.
And =\(\alpha ,\beta \) temperature co-efficient of resistance then
\({{R}_{T}}={{R}_{\circ }}\left( 1+\alpha t+\beta {{t}^{2}} \right)fort>{{300}^{\circ }}C\)
\({{R}_{T}}={{R}_{\circ }}\left( 1+\alpha t \right)fort<{{300}^{\circ }}C\)
\(\alpha =\frac{{{R}_{t}}-{{R}_{{}^\circ }}}{{{R}_{{}^\circ }}t}{}^\circ C\)
If \({{R}_{1}}\) and \({{R}_{2}}\) are the resistances at t1oC to t2oC respectively then \(\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1+\alpha {{t}_{1}}}{1+\alpha {{t}_{2}}}\)
\(\therefore \alpha =\frac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}{{t}_{2}}-{{R}_{2}}{{t}_{1}}}\)
The value of is different at different temperatures.
At a given temperature \(\alpha =\frac{1}{{{R}_{t}}}\left( \frac{dR}{dt} \right)\) at t0C.
Note : In general \({{R}_{0}}={{R}_{20}}\)
Graph shows the variation of resistivity with temperature for conductors,semiconductors and for alloys like managing and constantan. Since the resistivity of managing and constantan remains constant with respect to change in temperature, these materials are used for the bridge wires and resistance coils.
The resistivity of managing and constantan is almost independent of temperature.
Variation of resistance of some materials:
Strain dependence :
Just as the resistance of a conductor depends upon temperature, the resistance of a conductor depends upon strain By placing a conductor under tension (a form of stress that leads to strain in the form of stretching of the conductor), the length of the section of conductor under tension increases and its cross-sectional area decreases. Both these effects contribute to increasing the resistance of the strained section of conductor. Under compression (strain in the opposite direction), the resistance of the strained section of conductor decreases.
Light illumination dependence:
Some resistors, particularly those Made from semiconductors,exhibit photoconductivity, meaning that their resistance changes when light is shining on them. Therefore, they are called photoresistors (or light dependent resistors). These are a common type of light ditector.
1.From the following the quantity which is analogous to temperature in electricity is
Solution:
The quantity which is analogous to temperature in electricity is potential.
2.The resistance of a conductor is
1) Inversely proportional to the length.
2) Directly proportional to the square of the radius.
3) Inversely proportional to the square of the radius.
4) Directly proportional to the square root of the length.
Solution:
The resistance of a conductor is Inversely proportional to the square of the radius.
1.A wire 50 cm long and 1mm2 in cross-section carries a current of 4A when connected to a 2V battery. The resistivity of the wire is
Solution:
\(V=\frac{i\rho l}{A}\Rightarrow \rho =\frac{VA}{il}\)
\(1\times {{10}^{-6}}\,\,\Omega m\)
2.A wire of resistance 20 is bent in the form of a square. The resistance between the ends of diagonal is
Solution:
\({{R}^{1}}=\frac{20}{4}=5\)
\({{R}_{1}}=10\Omega ,\,{{R}_{2}}=10\)
\({{R}_{P}}=\frac{10}{2}=5\)
1.A uniform wire of resistance \(1\ \Omega /m\) having resistance is bent in the form of circle as shown in fig. If the equivalent resistance between M and N is\(1.8\ \Omega \) ,then the length of the shorter section is
Solution:
Let the resistance of shorter part MN be x.
Then resistance of longer part is (20 – x)\(\Omega \)
\({{R}_{eq}}=\frac{(20-x)x}{20-x+x}=1.8\ \Omega \)
Solving we get x =\(2\ \Omega \)
So length of shorter part = 2m
2.Twelve resistances each of resistance R are connected in the circuit as shown in fig. Net resistance between points A and Cwould be
Solution:
When a cell is connected across A and C, no current flows in the arms BG and ED due to symmetry in the arrangement. Then equivalent circuit will be as shown in Fig.(a) and (b). The effective resistance between A and C is
\(\frac{1}{{{R}_{eff}}}=\frac{1}{2R}+\frac{1}{3R}+\frac{1}{2R}\)
\(=\frac{3+2+3}{6R}=\frac{8}{6R}=\frac{4}{3R}\) or \({{R}_{eff}}=\frac{3R}{4}\)
1.A wire has a resistance R. What will be its resistance,
(a) if it is doubled on itself,
(b) if it is stretched so that
(i) its length is doubled
(ii) its radius is halved ?
Solution:
For a wire,
R = r \(\frac{L}{S}=\rho \frac{L}{\pi {{r}^{2}}}\)
If its dimensions are changed without changing its mass, its volume remains constant,
V = V¢ or SL = S¢L¢
(a) When the wire is doubled on itself, its length is halved and its area of cross-section is doubled.
S ´ L = S¢ ´ (L/2) or S¢ = 2S
\R¢ = \(\rho \frac{L'}{S'}=\rho \frac{\left( L/2 \right)}{\left( 2S \right)}=\frac{1}{4}\rho \frac{L}{S}=\frac{1}{4}R\)
(b) (i) When the length of the wire is doubled by stretching it,
S ´ L = 2L ´ S¢ or S¢ = S/2
\ R¢ = r \(\frac{L'}{S'}=\rho \frac{2L}{S/2}=\rho \frac{4L}{S}=4\rho \frac{L}{S}=4R\)
(ii) When the wire is stretched so as to reduce its radius to half (r¢ = r/2), we have
S¢ = pr¢2 = p (r/2)2 = pr2/4 = S/4
From S ´ L = S¢ ´ L¢ = (S/4) ´ L¢
Þ L¢ = 4L
\R = r\(\frac{{{L}^{'}}}{{{S}^{'}}}=\rho \frac{4L}{S/4}=16\rho \frac{L}{S}=16R\).
2.A metallic sphere of radius a is surrounded by a concentric thin metallic sheet of radius b. The space between these two electrodes is filled with a homogeneous poorly conducting material of resistivity r.
(a) Find the resistance between the two electrodes.
(b) Also derive an expression for the current density as a function of radius r, if the potential difference between the electrodes is V.
Solution:
(a) Consider a thin spherical layer between radii r and r + dr. The lines of current at all points of this layer are perpendicular to it. Therefore, such a layer can be treated as a conductor of length dr and cross-sectional area 4pr2. Thus,
dR = r \(\frac{dr}{4\pi {{r}^{2}}}\)
\R = \(\int\limits_{a}^{b}{dR}=\int\limits_{a}^{b}{\frac{\rho \text{ }dr}{4\pi {{r}^{2}}}}\)
\(=\frac{\rho }{4\pi }\left[ \frac{-\text{1}}{r} \right]_{a}^{b}=\frac{\rho }{4\pi }\left( \frac{1}{a}-\frac{1}{b} \right)\)
(b) The current density is given by
\(J=\frac{I}{S}=\frac{V}{RS}\)
At a radius r, S = 4pr2.
\ \(J\text{ }(r)=\frac{V}{4\pi {{r}^{2}}}\frac{4\pi }{\rho }\left( \frac{a\text{ }b}{b-a} \right)\)\(=\frac{V}{\rho {{r}^{2}}}\left( \frac{a\text{ }b}{b-a} \right)\).
This increases the conductivity of the material. The conductivity increases mean the resistivity decreases. Thus, when the temperature is increased in a semiconductor, the density of the charge carriers also increases and the resistivity decreases. ... The curve is non - linear for a wide range of temperature.
Resistivity:
Resistivity is the intrinsic property of the substance. It is independent of shape and
size of the body (i.e. l and A).
Resistivity depends on the temperature, for metals \({{\rho }_{t}}={{\rho }_{0}}\left( 1+\alpha t \right)\)Resistivity dependson the temperature, for metals i.e. resistivity increases with temperature.
\(\alpha =\frac{{{\rho }_{t}}-{{\rho }_{0}}}{{{\rho }_{0}}t}\) \(^{0}C\)
\(\alpha \)depends on temperature. At a given temperature \(\alpha =\frac{1}{{{\rho }_{t}}}\left( \frac{d\rho }{dt} \right)\) at t0C.
Variation of Resistivity and Resistance with Temperature:
Ø If \({{\rho }_{1}}\) is the resistivity of a material at temperature \({{t}_{1}}\) and \({{\rho }_{2}}\)is the resistivity of the same material at temperature \({{t}_{2}}\),then
\({{\rho }_{2}}={{\rho }_{1}}\left[ 1+\alpha \left( {{t}_{2}}-{{t}_{1}} \right) \right]\)
The constant factor\(\alpha \) is called the temperature coefficient of resistivity of the material. the reference temperature \({{t}_{1}}\)will be usually either \({{0}^{0}}\)C or \({{20}^{0}}C\)
Resistance of a conductor varies with temperature.
Variation of resistance of a conductor with temperature is given by
\({{R}_{2}}={{R}_{1}}\left[ 1+\alpha \left( {{t}_{2}}-{{t}_{1}} \right) \right]\)
where\(\alpha \) is called temperature co-efficient of resistance. Where \({{R}_{1}}\) is the resistance
of the conductor at a reference temperature \({{t}_{1}}\) (usuallyC or \({{20}^{0}}C\))
and \({{R}_{2}}\) is the resistance at temperature \({{t}_{2}}\).
\(\alpha =\frac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}{{t}_{2}}-{{R}_{2}}{{t}_{1}}}\) /\(^{0}C\)
resistivity of metals:
Electric resistance of the metallic wire increases when its temperature is raised.
Resistivity of Alloys:
With rise in temperature the resistivity of alloys also increases. But this increase is much smaller compared to pure metals. There are certain alloys such as manganin,constantan, nichrome, etc., whose resistivity is little affected by temperature, i.e., their
temperature coefficient of resistance is negligible.
Resistivity of Semiconductors :
The resistivity of semiconductors decreases with rise in temperature.
Resistivity of Electrolytes :
Ø With rise in temperature the resistivity of electrolytes also decreases. The reason is that with rise in temperature the viscosity of electrolytes decreases so that ions get more freedom to move inside the electrolytes. Hence the resistivity of the electrolytes decreases.
Ø Specific resistance or resistivity of a material is the resistance between opposite faces of a unit cube of the material.
Ø The specific resistance does not change with the shape of the resistor. It depends only on the material of the conductor at constant temperature.
Resistors can be joined together in many combinations. Any combination of resistors can be replaced by one equivalent resistance. When resistors are connected in series, the total resistance is greater than the individual resistances.
Example diagram:
Resistors in Series:
Two resistors are said to be in series if the same current passes through both resistors. For example, in the circuit shown, any current coming from the battery must pass through both resistors – the current has no place else to go.
At every instant of time all circuit elements connected in a series combination carry the same current.
The equivalent resistance of N resistors connected in series is given by
R = R1 + R2 + ….+ RN = \(\sum\limits_{i=1}^{N}{{{R}_{i}}}\)
The equivalent resistance of a series combination of resistances is always larger than any of the individual resistances.
Resistors in Parallel:
Two resistors are said to be in parallel combination if the current I splits into two currents I1 and I2 such that I = I1 + I2 but the voltage drop across each resistor is the same.
At every instant of time the potential difference across all circuit elements connected in parallel is the same.
For N resistors are connected in a parallel combination, the equivalent resistance is
\(\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+.......+\frac{1}{{{R}_{N}}}=\sum\limits_{i=1}^{N}{\frac{1}{{{R}_{i}}}}\)
The total resistance of any parallel combination of resistors must be less than any one of the individual resistors.
Resistor Colour codes:
Colour Number Multiplier Tolerance(%)
Black 0 1
Brown 1 \({{10}^{1}}\)
Red 2 \({{10}^{1}}\)
Orange 3 \({{10}^{1}}\)
Yellow 4 \({{10}^{1}}\)
Green 5 \({{10}^{1}}\)
Blue 6 \({{10}^{1}}\)
Violet 7 \({{10}^{1}}\)
Gray 8 \({{10}^{1}}\)
White 9 \({{10}^{1}}\)
Gold 5
Silver 10
No clour 20
The resistors have a set of co-axial coloured rings as shown above. The first two band from the end indicate the first two significant figures of the resistance in ohms. The third band indicates the decimal multilier (as listed in the above mentioned table). The
last band stands for tolerance or possible variation in percentage about the indicated values. Sometimes, this last band is absent and that indicates a tolerance of 20%. For example, if the four colours are orange, blue, yellow and gold, the resistance value i\(36\times {{10}^{4}}\Omega \) , with a tolerence value of 5%. To remember the sequence of colour code following sentence should keep in mind.
B.B.ROY of Great Britain having Very Good Wife with Gold and Silver Resistors in the higher range are made mostly from carbon. Carbon resistors are compact, inexpensive and thus find extensive use in electronic circuits.
Super Conductor:
Solution: The highest temperature is t1.
2.A piece of silver and another of silicon are heated from room temperature.The resistance of?
1) each of them increases
2) each of them decreases
3) Silver increases and Silicon decreases
4) Silver decreases and Silicon increases
Solution: Silver increases and Silicon decreases.
1.When two resistances are connected in parallel then the equivalent resistance is When one of the resistances is removed then the effective resistance is . The resistance of the wire removed will be
Solution:
\(\frac{{{R}_{1}}\,{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{6}{5}\,\,\) If \({{R}_{2}}\) is removed \({{R}_{1}}=2\Omega \)
\(\frac{2{{R}_{2}}}{2+{{R}_{2}}}=\frac{6}{5}\Rightarrow 5{{R}_{2}}=6+3{{R}_{2}}\Rightarrow {{R}_{2}}=3\Omega \)
2.The combined resistance of two conductors in series is 1. If the conductance of one conductor is 1.1 Siemen, the conductance of the other conductor in Siemen is
Solution:
R1 + R2 = 1 ohm.
\(\frac{1}{{{R}_{1}}}\)= 1.1 => R1 = \(\frac{10}{11}\)
R2 = 1 - R1
\(\therefore \frac{1}{{{R}_{2}}}=\frac{1}{1-{{R}_{1}}}\)
1.What is the resistance between A and B in figure.
Solution:
For the two resistors are in parallel, therefore,
\(\frac{1}{R}=\frac{1}{3}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}\) or R = 2W
This 2W resistor is in series with 8W, so
\({{R}_{AB}}=2\Omega +8\Omega =\underline{10\Omega }\)
2. For the network shown in figure, find the resistance between point a and point b.
Solution:
There are two parallel branches, each of resistance 3(20) = 60W.
Thus, \(1/R=\tfrac{1}{60}+\tfrac{1}{60}=\tfrac{2}{60},\)
and \(R=\underline{30\Omega }.\)
1.Three resistances of 12W, 16W and 20W are connected in parallel. What resistance must be connected in series with this combination to give a total resistance of 25W?
Solution:
The resistance R of the parallel combination is given by
\(\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}=\frac{1}{12}+\frac{1}{16}+\frac{1}{20}\)
\(=\frac{20}{240}+\frac{15}{240}+\frac{12}{240}=\frac{47}{240}\)
or R = 5.11 Then
\({{R}_{x}}+R=25\) or \({{R}_{x}}=25-5.11=\underline{19.89\Omega }\)
2.(a)At what temperature would the resistance of a copper conductor be double of its value at 0°C.
(b) Does this same temperature hold for all copper conductors, regardless of shape and size? (aCu = 4.0´10-3/C°)
(c) The resistance of a wire varies with temperature according to the relation
Solution:
R = R0(1 + aDq)
so \(\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{{{R}_{0}}\left[ 1+\alpha \left( {{T}_{2}}-0 \right) \right]}{{{R}_{0}}\left[ 1+\alpha \left( {{T}_{1}}-0 \right) \right]}=\frac{\left( 1+\alpha {{T}_{2}} \right)}{\left( 1+\alpha {{T}_{1}} \right)}\)
Here, R2 = 2R1 and t1 = 0
so 2 = 1 + aT2
or t2 = 1/a = 1/(4´10-3) = 250°C
(b) As here T2 = (1/a), does not include dimensions of the conductor (L and S), it is valid for all copper conductors whatever be their shape and size.