TB - UG

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 = ρ \frac{l}{A}

Where;

ρ =

resistivity of the element.

l =

length of the wire.

A =

area of cross section of the wire.
The electrical component rheostat, have 3 terminals,two terminals are the begining and end of the wire (

T_{1}

and

T_{2}

), while the 3rd terminal is the movable contact which can be moved along the length of the wire(

T_{3}

ρ

and

A

remains the same throughout; thus the resistance between terminals is decided by the length of wire between them.
The length

T_{1}

T_{3}

l_{1}

while

T_{3}

T_{2}

l_{2}

. This results in two resistances

R_{1}

and

R_{2}

; causing an equivalent circuit.By adding a voltage source across

T_{1}

and

T_{2}

; desired voltage can be obtained across

T_{3}

and ground by tuning the point of contact C on the rheostat.

example

Circuits involving electrical instruments

To convert a galvanometer into an ammeter a small resistance

S

(called the shunt) is connected in parallel with a galvanometer. A galvanometer has resistance

G

and full scale deflection current

i

. To convert this galvanometer into an ammeter of range

10 i

, a shunt

S = \frac{G}{9}

is connected in parallel with

G

. Now we want to measure the potential difference with the help of this ammeter. The maximum value of potential difference which can be measured with the help of this is:The voltage across galvanometer is:

V_{g} = i G

When the galvanometer is converted into an ammeter of range

10 i

by applying a shunt resistance

S = \frac{G}{9}

, the current is divided in inverse ratio of their resistance

\frac{1}{9} : 1

between shunt and ammeter .The current will be in ratio

9 : 1

.Current in shunt is

i = \frac{9}{10} \times 10 i = 9 i

The voltage across shunt resistance is:

V_{s} = 9 i (\frac{G}{9})

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

i G

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 resistances

S_{1}

and

S_{2}

connected in parallel. Find the condition for the bridge to be balanced.
Solution:The condition for Wheatstone bridge is:

\frac{P}{Q} = \frac{R}{S}

\Rightarrow S = \frac{S_{1} S_{2}}{S_{1} + S_{2}} [∵ S_{1}, S_{2}

in parallel]

\Rightarrow \frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

definition

Meter bridge

A meter bridge consists of a wire of length

1 m

and of uniform cross-sectional area stretched taut and clamped between two thick metallic strips bent at right angles with two gaps across which resistors are to be connected. The end points where the wire is clamped are connected to a cell through a key. One end of a galvanometer is connected to the metallic strip midway between the two gaps. The other end of the galvanometer is connected to a jockey which moves along the wire to make electrical connection.

R

is an unknown resistance connected across one of the gaps. Across the other gap, we connect a standard known resistance

S

. The jockey is connected to some point D on the wire, a distance

l_{1}

cm from the end A.The portion AD of the wire has a resistance

R_{c m} l_{1}

, where

R_{c m}

is the resistance of the wire per unit centimeter. The portion DC of the wire similarly has a resistance

R_{c m} (100 - l_{1})

.
The meter bridge works on the principle of Wheatstone bridge. At balance condition:

\frac{R}{S} = \frac{R_{c m} l_{1}}{R_{c m} (100 - l_{1})} = \frac{l_{1}}{100 - l_{1}}

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

and

3

ohm resistances respectively and are arranged to form a Wheatstone bridge.The resistance which the coil 'D' must be shunted in order that the bridge may be balanced is given by:Wheatstone condition:

\frac{R_{A}}{R_{C}} = \frac{R_{B}}{R_{D}}

R_{B} = R_{D}

R_{B} = \frac{3 \times R_{S}}{3 + R_{S}}

6 + 2 R_{S} = {3 R}_{S}

R_{S} = 6 Ω

example

Problems using meter bridge

When an unknown resistance and a resistance of

4 Ω

are connected in the left and right gaps of a meter bridge, the balance point is obtained at

50 c m

. The shift in the balance point, if a

4 Ω

resistance is now connected in parallel to the resistance in the right gap, isCase 1

\Rightarrow \frac{R}{4} = \frac{1}{1}

R = 4

\Rightarrow \frac{4}{2} = \frac{x}{100 - x}

200 = 3 x

x = \frac{200}{3}

Shift

= \frac{200}{3} - 50 = \frac{50}{3} = 16.7

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 difference

V_{0}

across AB is larger than the emf of either battery. One of the batteries is connected between the positive end A and the galvanometer. The jockey is adjusted to touch the wire at a point

C_{1}

so that there is no deflection in the galvanometer. The length

A C_{1} = l_{1}

is noted. Now, the first battery is replaced by the second and the length

A C_{2} = l_{2}

for the balance is noted. If

L

is the length

A B = L

, the emf of the first battery is,

E_{1} = \frac{l_{1}}{L} V_{0}

and that of the second battery is,

E_{2} = \frac{l_{2}}{L} V_{0}

Thus

\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}

definition

Internal resistance using potentiometer

We can also use a potentiometer to measure internal resistance of a cell. For this the cell (emf

E

) whose internal resistance (

r

) is to be determined is connected across a resistance box through a key

K_{2}

. With key

K_{2}

open, balance is obtained at length

l_{1}

(A N_{1})

. Then,

E = ϕ l_{1}

When key

K_{2}

is closed, the cell sends a current (

I

) through the resistance box (

R

). If

V

is the terminal potential difference of the cell and balance is obtained at length

l_{2} (A N_{2})

V = ϕ l_{2}

so,

\frac{E}{V} = \frac{l_{1}}{l_{2}} = \frac{I (r + R)}{I R}

\Rightarrow r = R (\frac{l_{1}}{l_{2}} - 1)

which is the internal resistance of the given cell.

law

Joule's Law of heating

According to Joule's Law of heating, heat generated in a conductor is given by:

H = V I t = \frac{V^{2}}{R} t = I^{2} R t

law

Joule's Law of heating

According to Joule's Law of heating, heat generated in a conductor is given by:

H = V I t = \frac{V^{2}}{R} t = I^{2} R t

definition

Electrical power

Electrical power is given by:

P = H / t = V I = I^{2} R

Its unit is

W

(Watt).

example

Example of a problem on Joule's Law

Example:
The internal resistance of a primary cell is

4

ohm. It generates a current of

0.2

ampere in an external resistance of

2

ohm. Find the rate at which chemical energy is consumed in providing the current.Solution:
The rate of consumption of chemical energy = Rate of generation of electrical energy.
The energy generated is calculated from the formula

P = V I

. Substituting

V = I R

(Ohm's law) in the above equation we get,

P o w e r = V I = I^{2} R

.

In this case, there are external and internal resistances. So, the equation is rewritten as

I^{2} (R + r)

.
Therefore, Chemical energy =

{0.2}^{2} (4 + 2) = 0.24 J / s .

Hence, the rate at which chemical energy is consumed in providing the current is

0.24 J / s

example

Power dissipated in resistors

Three equal resistors connected in series across a source of emf together dissipate

10 W

power. The power dissipated if the same resistors are connected in parallel is:Case 1

P = \frac{V^{2}}{3 R} = 10

Watt
Case 2

P = \frac{{3 V}^{2}}{R}

= 9 (\frac{V^{2}}{3 R})

= 90 W

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.

(P_{R})_{m a x} = \frac{ϵ^{2}}{4 r}

when R=r where

ϵ

is the emf of the cell and r is the internal resistance.
Example : Two identical batteries each of e.m.f. 2 V and internal resistance 1

Ω

are available to produce heat in an external resistance by passing a current through it. The maximum power that can be developed across R using these batteries isFor maximum current, the two batteries should be connected in series. The current will be maximum when external resistance is equal to the total internal resistance of cells i.e

R = r + r = 1 + 1 = 2 Ω

Thus,

E + E = I (R + r + r) \Rightarrow I = \frac{2 E}{R + 2 r} = \frac{2 \times 2}{2 + 2} = 1 A

Maximum power,

P = I^{2} R = 1^{2} \times 2 = 2 W

definition

Power supplied by a cell

Power supplied by a cell in a circuit is given as

P = V I = \frac{V^{2}}{r + R}

where

V

is the emf of the cell,

r

is the internal resistance and

R

is the load of the circuit.

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 emf

ε

and a resistance

R

.
Power in the emf source is

P_{s} = - V I

Power in the resistance is

P_{r}

By ohm's law,

V = I R

⟹ P_{r} = V I

∴ P_{s} + P_{r} = 0

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.

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.

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.

law

Law of intermediate temperatures

Let

ϵ_{θ 1, θ 2}

denote the thermo-emf of a given thermocouple when the temperatures of the junctions are maintained at

θ_{1}

and

θ_{2}

. Then,

ϵ_{θ 1, θ 2} = ϵ_{θ 1, θ 3} + ϵ_{θ 3, θ 2}

.
This is known as the law of intermediate temperatures

law

Law of intermediate metal

Suppose

ϵ_{A B}

=thermo-emf between metals A and B

ϵ_{B C}

=thermo-emf between metals B and C

ϵ_{A C}

=thermo-emf between metals A and C.
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,

ϵ_{A B} = ϵ_{A C} - ϵ_{B C}

.
This law is known as the law of intermediate metal

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

\dot{Q} = (Π_{A} - Π_{B}) I

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

i^{2} R t

and is called Thomson heat. The effect itself is called the Thomson effect.

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.

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.

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.

example

Example of power rating of a combination of bulbs

Example:
Two bulbs one of

200

volts,

60

watts & the other of

200

volts,

100

watts are connected in series to a

200

volt supply. Find the power consumed.Solution:
Here, two bulbs one of

200

volts,

60

watts & the other of

200

volts,

100

watts are connected in series to a

200

volt supply.
For bulb 1,

I = \frac{P}{V} = \frac{60}{200} = \frac{6}{20} A

Hence,

R = \frac{V}{I} = \frac{200 \times 20}{6} = \frac{4000}{6} Ω

Similarly, for bulb 2,

I = \frac{P}{V} = \frac{100}{200} = \frac{1}{2} A

Hence,

R = \frac{V}{I} = \frac{200 \times 2}{1} = 400 Ω

When bulbs are connected in series their resistances will added
up

R^{'} = (\frac{4000}{6}) + 400

R^{'} = \frac{6400}{6}

Hence, current is

I^{'} = \frac{V}{R^{'}} = \frac{200 \times 6}{6400} = \frac{12}{64} A

Hence,

P^{'} = V I = 200 \times \frac{12}{64}

P^{'} = 37.5 W

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

switch is closed :
We know that

P_{o} = \frac{V_{o}^{2}}{R}

.
Now, at

t = 0

, since voltage across

C = 0

, power consumed by

B_{1}

will be

P_{o}

.
Power consumed by

B_{2}

and

B_{3}

will be

\frac{P_{o}}{4}

each.
Hence, total power consumed at

t = 0

will be

2 (\frac{P_{o}}{4}) + P_{o} = \frac{3 P_{o}}{2}

.
After long time, power through

B_{1}

will be 0.
Hence total power will be

\frac{P_{0}}{2}

.
Brightness of

B_{2}

will be constant but that of

B_{1}

will decrease with time.

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 $$\dfrac{dQ}{dt}=C \dfrac{dV}{dt} \Rightarrow i_c=C \dfrac{dV_c}{dt}$$

definition

Voltage across a capacitor doesn't change suddenly

i = C \frac{d V}{d t}

An instantaneous change in the voltage across a capacitor would require taking limits.

lim_{t \to 0} i = C lim_{t \to 0} \frac{d V}{d t}

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.

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 = \frac{V_{s}}{R}

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 = \frac{E_{1}}{R_{1} + R_{2} + r_{1}}

\Rightarrow V_{C} - E_{2} = I R_{1}

V_{C} = E_{2} + \frac{R_{1} E_{1}}{R_{1} + R_{2} + r_{1}}

Energy stored

= \frac{1}{2} c v_{2}^{2}

= \frac{1}{2} c [E_{2} + \frac{R_{1} E_{1}}{R_{1} + R_{2} + r_{1}}]^{2}

example

RC circuits at the time of closing of switch

Only switch

S_{1}

closed in Figure. What is the charge on the

3 μ F

capacitor in

μ C

?Without power supply

V = 12

we can not solve.
When

S_{1}

is closed only , the capacitance

3 μ F

is charged to voltage

V = 12

.
Thus, the charge on

3 μ F

Q = 3 V = 3 \times 12 = 36 μ C

example

RC circuits at steady state

In the given circuit, the steady state voltage drop across the capacitor C isIn steady state

C_{^{'}}^{^{'}} = 0

V = I [r_{1} + r_{2}]

V_{C} = I r_{1}

= \frac{V r_{1}}{r_{1} + r_{2}}

formula

Discharging RC circuit

The differential equation of RC circuit is

\frac{d q}{q} = - \int \frac{1}{C R} d t

formula

Charging of an RC circuit

R \frac{d q}{d t} = \frac{ϵ C - q}{C}

where

ϵ

is the emf of the cell

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 = R C

where

τ

is the time constant and R and C are the values of resistance and capacitance respectively

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 (\infty) - (q (\infty) - q (0)) e^{- \frac{t}{R C}}

where:

q (t) :

Charge on the capacitor after time t

q (0) :

Initial charge on the capacitor

q (\infty) :

Charge on the capacitor at steady state

R C :

Time constant
Note:
The solution remains valid for other parameters of capacitors like voltage, current, etc.

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 (\infty) - (i (\infty) - i (0)) e^{- \frac{t}{R C}}

where:

i (t) :

Current in the resistor after time t

i (0) :

Initial current in the resistor

i (\infty) :

Current in the resistor at steady state

R C :

Time constant
Note:
The solution remains valid for other parameters of resistors like voltage.

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 = q_{0} (1 - e^{\frac{- t}{τ}})

where

q_{0}

is the initial charge of the capacitor and

q

is the charge at a time

t

formula

Voltage in a charging/discharging RC circuit

Charging

V (t) = V_{0} (1 - e^{- t / τ})

V_{r} (t) = V_{0} (e^{- t / τ})

Discharging

V_{c} (t) = V_{0} (e^{- t / τ})

V_{r} (t) = - V_{0} (e^{- t / τ})

where

τ = R C

is the time constant.

formula

Current in a charging/discharging RC circuit

Charging:

i = \frac{V}{R} e^{- t / τ}

Discharging

i = - \frac{V_{0}}{R} e^{- t / τ}

where

V_{0}

is the initial voltage.

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.

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

At steady state, capacitor will be fully charged and hence will have a potential of

V_{s}

across it.
Hence, charge stored in C in steady state is

q (\infty) = \frac{C}{V_{s}}

Charge stored in the capacitor after time t is given by:

q (t) = q (\infty) - [q (\infty) - q (0)] e^{- \frac{t}{R C}}

q (t) = \frac{C}{V_{s}} (1 - e^{- \frac{t}{R C}})

Energy stored in capacitor after time t is given by:

E (t) = \frac{q^{2} (t)}{2 C}

E (t) = \frac{\frac{C}{V_{s}} (1 - e^{- \frac{t}{R C}})}{2 C}

formula

Heat generated in resistor in RC circuit

A capacitor

C

is charged to

V_{0}

volts, a resistor

R

is connected across it. The heat produced in the resistor is

C V_{0}^{2} / 2

formula

Time to reach specific percentage of charge in a RC circuit

\frac{q}{q_{0}} = 1 - e^{- \frac{t}{τ}}

l n (1 - \frac{q}{q_{0}}) = - \frac{t}{τ}

t = τ l n (\frac{1}{1 - \frac{q}{q_{0}}})

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.

example

Solving complicated RC circuit excited by DC

Example:
Consider the circuit shown in the figure. The capacitor is initially uncharged. Find the current across

R_{2}

after time t.
Solution:
To find the time-constant, the emf source is short-circuited. The equivalent resistance is then given by:

R_{e q} = (R_{1} + R_{3}) | | R_{2}

R_{e q} = \frac{5000}{11} Ω

Time-constant is then given by:

τ = R_{e q} C = \frac{1}{22} s

Now, voltage across the capacitor initially is

V (0) = 0

and remains the same at the time of switching.
This is the same as the voltage across

R_{2}

.
By ohm's law,

i_{2} (0) = 0

In steady state, the capacitor is fully charged and acts as open circuit. Then current across

R_{2}

can be found using KCL.

i_{2} (\infty) = \frac{20}{R_{1} + R_{2} + R_{3}} = \frac{1}{275} A

Using general solution,

i_{2} (t) = i_{2} (\infty) - [i_{2} (\infty) - i_{2} (0)] e^{- \frac{t}{τ}}

i_{2} (t) = \frac{1}{275} (1 - e^{- 22 t}) A

Note:
This approach is useful when the resistors and capacitors network can be separated into two groups.

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 be

i = \frac{20 + 10}{4 + 2} = 5 A

.
Reading of voltmeter just after closing the switch will be

10 - 2 i = 0 V

.
After long time, entire voltage will appear across the capacitor, hence, Q=CV=2X30=60C

Monday, July 31, 2017

Current Electricity2

Rheostat as a potential divider

Circuits involving electrical instruments

Solving circuit of Wheatstone bridge

Meter bridge

Null point

Problems using meter bridge

Comparision of emfs using a potentiometer

Internal resistance using potentiometer

Joule's Law of heating

Joule's Law of heating

Electrical power

Example of a problem on Joule's Law

Power dissipated in resistors

Maximum power transfer theorem

Power supplied by a cell

Energy conservation in a circuit

Seebeck effect

Neutral and inversion temperature

Working of a thermocouple

Law of intermediate temperatures

Law of intermediate metal

Peltier effect

Thomson effect

Voltage rating of a bulb

Power rating of a bulb

Difference between breaking voltage and voltage rating of a bulb

Example of power rating of a combination of bulbs

Bulb brightness and power dissipated

Relationship between current and voltage of a capacitor

Voltage across a capacitor doesn't change suddenly

Behaviour of capacitor on closing the switch

Capacitor as an open circuit

RC circuits at the time of closing of switch

RC circuits at steady state

Discharging RC circuit

Charging of an RC circuit

Time constant in an RC circuit

General solution of RC circuit for a capacitor

General solution of RC circuit for a resistor

Charge on a capacitor in a charging RC circuit

Voltage in a charging/discharging RC circuit

Current in a charging/discharging RC circuit

Non-ideal capacitor

Energy stored in a capacitor in RC circuit

Heat generated in resistor in RC circuit

Time to reach specific percentage of charge in a RC circuit

Sudden change in voltage s forced across a capacitor

Solving complicated RC circuit excited by DC

Charge on a capacitor in a charging/discharging circuit