Rheostat as a potential divider
A rheostat is a resistance element with considerable length. It is usually metallic in nature. The defining equation is R=ρlA
Where;
The electrical component rheostat, have 3 terminals,two terminals are the begining and end of the wire (
The length
and ground by tuning the point of contact C on the rheostat.
52
example
Circuits involving electrical instruments
To convert a galvanometer into an ammeter a small resistance S
When the galvanometer is converted into an ammeter of range
The voltage across shunt resistance is:
Since the galvanometer(ammeter) and shunt are in parallel voltage across them is same, hence the maximum value of potential difference which can be measured with the help of this ammeter is
.
53
example
Solving circuit of Wheatstone bridge
Example:
In a Wheatstone's bridge, three resistances P,Q and R are connected in three arms and the fourth arm is formed by two resistancesS1
In a Wheatstone's bridge, three resistances P,Q and R are connected in three arms and the fourth arm is formed by two resistances
Solution:The condition for Wheatstone bridge is:
54
definition
Meter bridge
A meter bridge consists of a wire of length 1 m
The meter bridge works on the principle of Wheatstone bridge. At balance condition:
55
definition
Null point
The
null point or balanced point is an arrangement of resistors across the
arms of a Wheatstone bridge or meter bridge such that the deflection in
the galvanometer is zero. A,B,C and D are four coils of wires of 2,2,2
56
example
Problems using meter bridge
When an unknown resistance and a resistance of 4Ω
Shift
cm
57
definition
Comparision of emfs using a potentiometer
Comparison of Emfs:
The driving circuit of a potentiometer is set up with a strong battery so that the potential differenceV0
The driving circuit of a potentiometer is set up with a strong battery so that the potential difference
and that of the second battery is,
Thus
58
definition
Internal resistance using potentiometer
We can also use a potentiometer to measure internal resistance of a cell. For this the cell (emf E
When key
so,
which is the internal resistance of the given cell.
59
law
Joule's Law of heating
According to Joule's Law of heating, heat generated in a conductor is given by:
H=VIt=V2Rt=I2Rt
60
law
Joule's Law of heating
According to Joule's Law of heating, heat generated in a conductor is given by:
H=VIt=V2Rt=I2Rt
61
definition
Electrical power
Electrical power is given by: P=H/t=VI=I2R
Its unit is
(Watt).
62
example
Example of a problem on Joule's Law
Example:
The internal resistance of a primary cell is4
The internal resistance of a primary cell is
The rate of consumption of chemical energy = Rate of generation of electrical energy.
The energy generated is calculated from the formula
In this case, there are external and internal resistances. So, the equation is rewritten as
Therefore, Chemical energy =
Hence, the rate at which chemical energy is consumed in providing the current is
.
63
example
Power dissipated in resistors
Three equal resistors connected in series across a source of emf together dissipate 10 W
Case 2
64
definition
Maximum power transfer theorem
The
maximum power transfer theorem states that, to obtain maximum external
power from a source with a finite internal resistance, the resistance of
the load must equal the resistance of the source as viewed from its
output terminals.
(PR)max=ϵ24r
when R=r where
Example : Two identical batteries each of e.m.f. 2 V and internal resistance 1
Thus,
Maximum power,
65
definition
Power supplied by a cell
Power supplied by a cell in a circuit is given as
P=VI=V2r+R
where
is the load of the circuit.
66
definition
Energy conservation in a circuit
In an electrical circuit, total power supplied equals the total power delivered at all instants.
Consider a circuit consisting of an ideal voltage source with emfε
Consider a circuit consisting of an ideal voltage source with emf
Power in the emf source is
Power in the resistance is
By ohm's law,
67
definition
Seebeck effect
Two
metallic strips made of different metals are joined at the ends to form
a loop. If the junctions are kept at different temperatures, there is
an electric current in the loop. This effect is called the Seebeck
effect and the emf developed is called the Seebeck emf or thermo-emf.
68
definition
Neutral and inversion temperature
The temperature of the hot junction at which the thermo-emf is maximum is called the neutral temperature and the temperature at which the emf changes its sign (current reverses) is called the inversion temperature.
69
definition
Working of a thermocouple
The working principle of thermocouple is based on three effects - Seebeck Effect, Peltier Effect and Thomson Effect.It
comprises of two dissimilar metals, A and B, joined together to form
two junctions, p and q, which are maintained at the temperatures T1 and
T2 respectively which generates the Peltier emf within the circuit and
it is the function of the temperatures of two junctions.The total emf or
the current flowing through the circuit can be measured easily by the
suitable device.Now, the temperature of the reference junctions is
already known, while the temperature of measuring junction is unknown.
The output obtained from the thermocouple circuit is calibrated directly
against the unknown temperature. Thus the voltage or current output
obtained from thermocouple circuit gives the value of unknown
temperature directly.
70
law
Law of intermediate temperatures
Let ϵθ1,θ2
.
This is known as the law of intermediate temperatures
This is known as the law of intermediate temperatures
71
law
Law of intermediate metal
Suppose ϵAB
Also, suppose the temperatures of the cold junctions are the same in the three cases and the temperatures of the hot junctions are also the same in the three cases. Then,
.
This law is known as the law of intermediate metal
This law is known as the law of intermediate metal
72
definition
Peltier effect
The Peltier effect is the presence of heating or cooling at an electrified junction of two different conductors. When
a current is made to flow through a junction between two conductors, A
and B, heat may be generated or removed at the junction. The Peltier
heat generated at the junction per unit time,\dot { Q } , is equal to
Q˙=(ΠA−ΠB)I
73
definition
Thomson effect
If
a metallic wire has a nonuniform temperature and a current is passed
through it, heat may be absorbed or produced in different sections of
the wire. This heat is over and above the Joule heat i2Rt
and is called Thomson heat. The effect itself is called the Thomson effect.
74
definition
Voltage rating of a bulb
The
voltage rating of an incandescent bulb is the voltage that bulb is
designed to operate in the circuit measured at the base of the bulb and
in which published data watts, amps, lumens, colour temperature and life
hours are measured.
75
definition
Power rating of a bulb
The
power rating on a light bulb indicates how much power it would
dissipate when it is hooked up to the standard household voltage of
120V.
76
definition
Difference between breaking voltage and voltage rating of a bulb
Voltage rating of a bulb is the voltage at which the bulb operates.
If an excess voltage is applied to the bulb, a dangerously high electric field is induced which may damage the bulb. The minimum value of voltage at which breakdown occurs is called the breakdown voltage of the bulb.
If an excess voltage is applied to the bulb, a dangerously high electric field is induced which may damage the bulb. The minimum value of voltage at which breakdown occurs is called the breakdown voltage of the bulb.
77
example
Example of power rating of a combination of bulbs
Example:
Two bulbs one of200
Two bulbs one of
Here, two bulbs one of
For bulb 1,
Hence,
Similarly, for bulb 2,
Hence,
When bulbs are connected in series their resistances will added
up
Hence, current is
Hence,
78
definition
Bulb brightness and power dissipated
An unchanged capacitor is connected in circuit as shown in figure Power ratings of bulbs are given in diagram At t=0
We know that
Now, at
Power consumed by
Hence, total power consumed at
After long time, power through
Hence total power will be
Brightness of
will decrease with time.
79
definition
Relationship between current and voltage of a capacitor
The relationship between a capacitors voltage and current define its capacitance and its power. We have Q=CV.
Then $$\dfrac{dQ}{dt}=C \dfrac{dV}{dt} \Rightarrow i_c=C \dfrac{dV_c}{dt}$$
Then $$\dfrac{dQ}{dt}=C \dfrac{dV}{dt} \Rightarrow i_c=C \dfrac{dV_c}{dt}$$
80
definition
Voltage across a capacitor doesn't change suddenly
An instantaneous change in the voltage across a capacitor would require taking limits.
i.e. the rate of change of the voltage (dv/dt) be infinite, and hence the current would have to be infinite which is not possible.
81
definition
Behaviour of capacitor on closing the switch
Voltage
across a capacitor does not change suddenly. Hence, for an uncharged
capacitor, just after the switch is closed voltage remains zero. This is
analogous to a short circuit. Hence, an initially uncharged capacitor
behaves as a short circuit just after closing the switch.
Example:
Find the current in the circuit shown in the attached figure just after the closing of the switch. The capacitor is initially uncharged.
Solution:
Just after the closing of the switch, capacitor behaves as a short circuit. Hence, the current is given by:
I=VsR
Example:
Find the current in the circuit shown in the attached figure just after the closing of the switch. The capacitor is initially uncharged.
Solution:
Just after the closing of the switch, capacitor behaves as a short circuit. Hence, the current is given by:
82
example
Capacitor as an open circuit
In the steady state, the energy stored in the capacitor is :In steady state current flow in capacitor branch is zero.
I=E1R1+R2+r1
Energy stored
83
example
RC circuits at the time of closing of switch
Only switch S1
When
Thus, the charge on
84
example
RC circuits at steady state
In the given circuit, the steady state voltage drop across the capacitor C isIn steady state C′′=0
85
formula
Discharging RC circuit
The differential equation of RC circuit is
dqq=−∫1CRdt
86
formula
Charging of an RC circuit
is the emf of the cell
87
definition
Time constant in an RC circuit
The
time constant of an RC circuit is the time required to charge the
capacitor, through the resistor, by 63.2 percent of the difference
between the initial value and final value or discharge the capacitor to
36.8 percent.
τ=RC
is the time constant and R and C are the values of resistance and capacitance respectively
88
formula
General solution of RC circuit for a capacitor
General solution of RC circuit for a capacitor after time t is given by :
q(t)=q(∞)−(q(∞)−q(0))e−tRC
where:
Time constant
Note:
The solution remains valid for other parameters of capacitors like voltage, current, etc.
Note:
The solution remains valid for other parameters of capacitors like voltage, current, etc.
89
definition
General solution of RC circuit for a resistor
General solution of RC circuit for a resistor after time t is given by :
i(t)=i(∞)−(i(∞)−i(0))e−tRC
where:
Time constant
Note:
The solution remains valid for other parameters of resistors like voltage.
Note:
The solution remains valid for other parameters of resistors like voltage.
90
formula
Charge on a capacitor in a charging RC circuit
Charge on a capacitor in a charging circuit is given by the following equation.
q=q0(1−e−tτ)
where
.
91
formula
Voltage in a charging/discharging RC circuit
Charging V(t)=V0(1−e−t/τ)
Discharging
where
is the time constant.
92
formula
Current in a charging/discharging RC circuit
Charging: i=VRe−t/τ
Discharging
where
is the initial voltage.
93
definition
Non-ideal capacitor
If
the dielectric material between the plates of a capacitor has finite
resistivity as opposed to infinite resistivity in case of an ideal
capacitor then there will be a small amount of current flowing between
the plates of the capacitor.Lead resistance and plate effects also exist
in non-ideal capacitor.
94
example
Energy stored in a capacitor in RC circuit
Example:
Consider the circuit shown in the attached figure. Find the energy stored in the capacitor after time t. The initial charge on the capacitor is 0.
Solution:
Given,q(0)=0
Consider the circuit shown in the attached figure. Find the energy stored in the capacitor after time t. The initial charge on the capacitor is 0.
Solution:
Given,
At steady state, capacitor will be fully charged and hence will have a potential of
Hence, charge stored in C in steady state is
Charge stored in the capacitor after time t is given by:
Energy stored in capacitor after time t is given by:
95
formula
Heat generated in resistor in RC circuit
A capacitor C
96
formula
Time to reach specific percentage of charge in a RC circuit
97
definition
Sudden change in voltage s forced across a capacitor
If
a source of voltage is suddenly applied to an uncharged capacitor, the
capacitor will draw current from that source, until the capacitors
voltage equals that of the source. Once the capacitor voltage reached
this charged state, its current decays to zero.
98
example
Solving complicated RC circuit excited by DC
Example:
Consider the circuit shown in the figure. The capacitor is initially uncharged. Find the current acrossR2
Consider the circuit shown in the figure. The capacitor is initially uncharged. Find the current across
Solution:
To find the time-constant, the emf source is short-circuited. The equivalent resistance is then given by:
Time-constant is then given by:
Now, voltage across the capacitor initially is
This is the same as the voltage across
By ohm's law,
In steady state, the capacitor is fully charged and acts as open circuit. Then current across
Using general solution,
Note:
This approach is useful when the resistors and capacitors network can be separated into two groups.
99
example
Charge on a capacitor in a charging/discharging circuit
Initially
capacitor was uncharged and switch was open. Switch is closed at t=0.
Ammeter and voltmeter are ideal. [All units are in SI]After
long time, current through the circuit will be zero since capacitor will
behave as open circuit. Therefore voltmeter will read 10V.
Just after closing the switch, voltage across capacitor will be 0V. Hence current will bei=20+104+2=5A
Just after closing the switch, voltage across capacitor will be 0V. Hence current will be
Reading of voltmeter just after closing the switch will be
After long time, entire voltage will appear across the capacitor, hence, Q=CV=2X30=60C
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