Resistance - thetechnoserverthetechnoserver.weebly.com/uploads/2/0/7/5/20757078/__current... · Internal Resistance: Every source of emf has a resistance of its own, known as the

Resistance

Resistance in Series:

“The Connected resistances are said to be in series when a potential

difference that is applied across the combination is the sum of the resulting

potential differences across the individual resistances".

The current through each resistor in series is necessarily the same but this is

not the sufficient condition for resistors in series. The condition becomes

sufficient when the current through each resistor is the same as the current

through the terminals of the network.

A combination of three resistances R1, R2 and R3 in series. The current through

each resistor is the same as the current i through the terminals A or B of the

networks. Suppose that V1, V2 and V3 are the potential differences across R1' R2

and R3 respectively. If V = VA - VB be the potential difference across the network,

then according to the definition of series combination.

V = V1 + V2 + V3

== iR1 + iR2 + iR3

== i (R1 + R2 + R3)

Important Points:

Total potential (V) is divided in such a way that :

V1 : V2 : V3 == R1 : R2 : R3

Or,

If R1= R2 =R3 then V1= V2 = V3= V/3 and Req =3R

Similarly, if there are n resistors in series,

Req=R1 +R2+ .. + Rn

If R1= R2 =R3' then V1= V2 = V3= V/3 and Req =3R

, Similarly, if there are n resistors in series,

Req=R1 +R2+ .. + Rn

RESISTANCES IN PARALLEL

The connected resistances are said to be in parallel when a potential

difference that is applied across the combination is the same as' the

resulting potential difference across the individual resistances. It shows

three resistances in parallel V = potential difference across each resistor =

potential difference across the combination between A and B

The currents in three resistances are:

i1 =R1V

, i2 =R2V

and i3 =R3V

The total current is

Or, Req

V =R1

V +R2

V +R3

V

Or, Req

1 =R1

1 +R2

1 +R3

1

For n resistors in parallel,

Req

1 =R1

1 +R2

1 +...+Rn

1

For two resistors in parallel,

Req

1 =R1 + R2

R1 R2

(i) here equivalent resistance is less than each of the individual

resistance.

(ii) i1 : i2 : i3 =R1

1:

R2

1:

R3

1

Example. What is the equivalent resistance between two points A and B?

Solution:

Taking R2 and R3 in parallel, the resulting will be as following.

Now, equivalent resistance will be in series combination.

. . Req = R1 + R + R4

= (2 + 2+ 2) Ω = =6Ω

ELECTRIC CURRENT AND KIRCHHOFF'S LAW

Internal Resistance: Every source of emf has a resistance of its own, known

as the internal resistance (r). For example, in case of a cell, the components

inside it offer a resistance to the flow of current. The value of the internal

resistance of a cell depends on: (a) the surface area of its electrodes, (b) the

separation between them and (c) the nature, concentration and temperature

of its electrolyte.

A simple electric circuit

When a current i flows through C, the source C does work to transfer positive

charge from one end A to the other end B. lf E is the emf of the cell and r its

internal resistance, then a part v of E is used up in transferring

charge from A to B. From Ohm's law, this part is given by

v = ir ... (i)

The remaining part V = (E - v) drives the charge through the external resistor

R; this part is given by

V = iR ... (ii)

I

Now, E = V + v ... (iii)

Equations (i) to (Hi) give

V =(R + r)

ER"

EV =

R + rR

Also, i =R

E - V =r

E - V..... (iv)

or r =i

E - V =V

(E - V)R

E - V = ir ... (v)

Notice from Equation (iv) that if i = 0 E = V, i.e., if no current is drawn from the ell

(i.e., in an open circuit), the potential difference V across its terminals equals its

emf E. If ≠ 0, V is less than E.

From (v), E=V, when r = O, V < E. lf r ≠O

Kirchhoff's Rules: Kirchhoff found two rules for determining the current and

resistance in a complicated circuit. These laws are as follows:

1. In an -electric circuit-the algebraic sum of the currents meeting at any

junction in the circuit is zero, that is ∑I = 0

When applying this law, the current going towards the junction is taken as

positive while that going away is negative. In the following (a) and (b)

11 + 13 - 12 - 14 -Is = 0 or 11 + 13 = 12 + 14 + Is

Thus the sum of the currents flowing toward the junction is equal to the sum of the

currents flowing away, from the junction. Therefore, when a steady current flows

in a circuit then there is neither any accumulation of charge at any point in the

circuit nor any depletion of charge from there. Thus, Kirchhoff's first rule expresses

the conservation of charge.

2. In any closed mesh of a circuit, the sum of the products of the current and

the resistance in each part of the mesh is equal to the algebraic sum of the

e.m.f.'s in that mesh, that is, ∑(IR) = ∑E

In applying this law, when we traverse in the direction of current then the

product of the current and the corresponding resistance is taken as

positive, and the e.m.f. is taken as positive when we traverse from the

negative to the positive electrode of the cell through the electrolyte.

Applying Kirchhoff's" rule to mesh 1, (or closed mesh ABFGA)

l1R1 - l2R2 = E1 - E2

Similarly, for the mesh 2, (or closed mesh BCDFB) we have

l2R2 + (l1 + l2)R3 = E2

From these equations, we can determine the values of currents in different

parts of the circuit.

Grouping of Cells: A cell is a source of electric current. A single cell cannot give a

strong current.

Therefore, to get strong current two or more cells are to be combined. The

combination of cells is called a 'battery'. Cells can be combined in three

ways: (a) In series, (b) In parallel and (c) In mixed grouping.

1. In series: In this combination, the negative pole of the first cell is

connected to the positive pole of the second cell, the negative pole of the

second. to the positive pole of the third, the negative pole of the third to the

positive pole of the fourth, and so on

Suppose, n cells each of e.m.f E and internal resistance r are connected in

series. These cells are sending current in an external resistance R. Then,

Total e.m.f of the cells = n E

Total internal resistance = n r

Therefore, Total resistance of the circuit = (nr + R)

If the current in the circuit be l, then = I =(nr + R)

nE

Important Points:

1. If r« R, then from eq. (i) I =R

nE

(approx,) that is, if the internal resistance of the connected

cells is much smaller than the external resistance, then the current

given by these cells will be nearly n times the current given by one

cell. Hence, when the internal resistance often connected cells is

much smaller than the external resistance, then the 'cells should be

connected

in series to obtain it strong current.

2. If nr » R, then, I =R

nE =rE (approx), that is, if internal resistance

of the connected cells is much greater than the external resistance,

then nearly the same current is obtained by single cell. Hence, there

is no advantage of connecting cells in series.

3. If in series grouping of n cells, n1 cells are reversed then

Eeq = (n- n1)E - n1E = (n - 2n1)E and req = nr

So I = (n - 2n1) E

R + nr

2. In Parallel: In this combination, the positive poles of all the cells are

connected to one point, and the negative poles to another point . Suppose

m cells, each of e.m.f. E and internal resistance r, are connected in parallel

and this battery of m cells is connected to an external resistance R. Since

the

cell are connected in parallel, the e.m.f. of the battery will also be E. If the

equivalent internal resistance of the cell be R1, then

R1

1 =r1 +

r1 + ..... upto m terms=

rm

or R1 =mr

Therefore, Total resistance of the circuit = (r/m + R). If the current in the

external circuit be l,

Then, l =(r/m)+ R

E =r + mR

mE

Important Points:

i. If r<<R, that is, if the internal resistance of the cells is much

smaller than the external resistance, then r can be

neglected in comparison to mR. Then, from eq. (i), l = E/R

(approx.), i.e., total current will be equal to the current

given by a single cell. Hence, there is no advantage of

connecting the cells of small internal resistance in

parallel.

ii. If (r/m)>>R, that is, if the internal resistance of the cells in

larger than the external resistance, then the current will

be I =R

nE (approx.). This current is nearly m times the

current given by r

a single cell. Hence, when the internal resistance of the

cell is much larger than the external resistance, then the

cells should be connected in parallel.

Mixed Grouping: - In this combination, a certain number of cells are

connected in series, and all such series. Combinations are then connected

mutually in parallel.

Suppose n cells are connected in each series, and such m rows are connected in

parallel. Let the e.m.f. of each cell be E and the internal resistance be r. This

battery of cells is sending current in an external resistance R.

The total e.m.f. of n cells connected in one series is nE. Since all the series are

connected in parallel the e.m.f. of the battery as a whole will also be nE.

Similarly, the total internal resistance of cells in a series is nr. Such m rows are

connected in parallel. Hence,if the internal resistance of the whole battery be R1

then

R1

1 =nr1 +

nr1 +... upto m terms=

nrm

or R1 =mnr

Therefore, total resistance of the circuit = mnr` j+ R$ . . If the current in the

external circuit be l, then

I =nr +mR

mnE …… (i)

In eq. (i) for the value of l to be maximum, the value of (nr + mR)

should be minimum. Now

Nr +mR = { √nr - √mR }2 + 2√(mnRr)}

Therefore, for (nr +mR) to be minimum, {√nr - √mR} = 0

Or √nr = √mR or nr = mR or R = nr

m

But, nr/m is the internal resistance of the whole battery. Thus, in mixed grouping

the current in the external Circuit will be maximum when the internal

resistance of the battery is equal to external resistance. By substituting nr/m

= R in eq. (i), we can see that the maximum current in the external circuit will

be nE/2R or mE/2r.

Moving Coil Galvanometer

Moving coil galvanometer is a device used to detect small current flowing in an electric

circuit. With suitable modifications, it can be used to measure current and potential

difference.

Conversion of galvanometer into an Ammeter

An ammeter is an instrument which is used to measure current in a circuit in ampere (or

miIli-ampere or micro- ampere). Hence, it is always connected in series in the circuit. Since,

the galvanometer coil has some resistance of its own, therefore, to convert a galvanometer

into an ammeter, its resistance is to be decreased so, to convert a galvanometer into

ammeter a low resistance, called shunt (S) is connected in parallel to the galvanometer as

shown in .

Here, ig = S + G

Si

And RA = (G + S)

S^ h G^ h

Where RA = Resistance of ammeter

S = Resistance of Stunt

G = Resistance of Galvanometer

Conversion of Galvanometer into Voltameter

A voltameter is an instrument which is used to measure the potential difference

between two points of an electric circuit directly in volt (or milli-volt or

micro-volt). Hence, it is connected in parallel across is to be measured. When it is

connected. Since, the resistance of coil of galvanometer of its own is low, hence, to

convert a galvanometer into a voltmeter, high resistance R in series is connected with the

galvanometer.

Here, ig =R +G

V

Where R + G = Rv = resistance of voltameter

Example. A millimeter of range 10 m A and resistance 9 X is joined in a circuit as

shown. The metre gives full – scale deflection for current I when A and B are used

as its terminals i.e., current enters at A and leaves at B (C is left isolated). The

value of I is

(A)100mA (B) 900mA (C) 1A (D) 1.1A

Solution. According to loop rule,

-9X10-0.9X10+0.1X(I-10) = 0

Or, I – 10 = 0.1

90 + 9 = 990

Therefore, I = 1000mA = 1A

Example. In the circuit shown in reading of votameter is V1 when only S1 is

closed, reading of voltmeter is V 2 when only S2 is closed. The reading

of voltmeter is V3 when both Sl and S2 are closed then

(A)V2>V1>V3 (B)V3>V2>V1 (C)V3>V1>V2 (D)V1>V2>V3

Solution. When S1 is closed,

I =4Rf

v1 = I X 3R =4Rf

X 3R

v1 =43f

When S2 is closed,

I =7Rf

v2 = I X 6R =7Rf

X 6R

v2 =76f

When S1 and S2 both are closed,

I =3Rf

v3 = I X 2R =3Rf

X 2R

v3 =32f

Example: In the circuit shown in the reading of ammeter is the same with both

switches open as with both closed. Then find the resistance R. (ammeter is ideal)

Solution.

Step-I: Discuss the circuit when both switches open:

According to loop rule: 1.5 - 300 I-100 I - 50 I = 0

Therefore, I =4501.5 =

450015 =

3001

A

Step – II: Discuss the circuit after closing the switch.

In loop ABCDEA,

-IR + 1.5 – 300I1=0

Or 300I1 + IR = 1.5

In loop BCGFB,

-100I + (I1 – I)R =0

Or (I1 – I)R = 100I

I1R = (100+R)I

Therefore, I1 = R

100 + R^ hI

From eqn (1) & (2)

300R

(100 + R)I + IR =1.5

300R

(100 + R)3001 +

300R =1.5

R=600X

Example. The battery in the diagram is to be charged by the generator G The

generator has a terminal voltage of 120 volts when the charging current is10

amperes. The battery has an emf of 100 volts and an internal resistance of 1 ohm.

In order to charge the battery at 10 amperes charging current, the resistance R

should be set at:

(A) 0.1X (B) 0.5 X (C) 1.0 X

Solution. VA - VB = - { algebraic sum of rise up and drop up of voltage }

120 = -{-IR-1 X I -100}

120 = IR + I +100

Or 20 = 10R + 10

Therefore, R = 1 X

Example. The resistance of thin silver wire is 1.0X at 20°C. The wire is placed in a

liquid bath and its resistance rises to 1.2X . What is the temperature of the bath?

α for silver is 3.8X10-3 per °C.

Solution. R(T) = R0 *1+α(T-T0)]

Hence, R(T) = 1.2X , R0 = 1.0X

Α = 3.8 X 10-3 per°C and T0 = 20°C

Substituting the values, we have

1.2 = 1.0[1+3.8 X 10-3(T-20)]

Or 3.8 X 10-3 (T-20) = 0.2

Solving this, we get T = 72.6°C

Example. A resistance R of thermal coefficient of resistivity = α is connected in

parallel with a resistance = 3R, having thermal coefficient of resitivity = 2α. Find

the value of αeff.

Solution. The equivalent resistance at 0°C is

R0 = R10 + R20

R10 R20 ….(1)

The equivalent resistance at t°C is

R = R1 + R2

R1R2 …(2)

But, R1 = R10(1+αT) …(3)

R2 = R20(1+2αT) …(4)

And R = R0(1 + αefft) …(5)

Putting the value of (1), (3), (4) and (5) in eqn (2)

αeff = 45 a

Example. (a) The current density across a cylindrical conductor of radius R varies according

to the equation

J = J0(1 -Rr

) , where r is the distance from the axis. Thus the current density is a

maximum Jo at the axis r = 0 and decreases linearly to zero at the surface r = R

Calculate the current in terms of Jo and the conductor's cross sectional area is

A = nR 2

(b) Suppose that instead the current density is a maximum J 0 at the surface and

decreases linearly to zero at the axis so that J = J = J0 Rr . Calculate the current.

Solution. (a) We consider a hollow cylinder of radius r and thickness dr.

The cross-sectional area of considered element is dA = 2πrdr

The current in considered element is

dI = JdA = J0 (1 -Rr

)2rrdr

or dI = 2rJ0(1-Rr

)rdr

Therefore, I = 2rJ0 (0

R

# 1-Rr

)rdr

I = I = J0 3rR2

= J0 3A

(b) I =R

2rJ0

0

R

#Rr2

dr

I =R

2rJ0

Rr3; E

0

R

I =R

2rJ0

Rr3

=3

2AJ0

Example. A network of resistance is constructed with Rl & R2 as shown in the . The

potential at the points 1, 2,3,……….., N are V1, V2, V3,…….., VN respectively each

having a potential k time smaller than previous one. Find:

1) R2

R1 andR3

R2 in terms of K

2) Current that passes through the resistance R2 nearest to the V0 in terms V0,

k and R3.

Solution.

1) According to kcL,

I = I1 + I2

Or R1

V0 -kV0

=R2

kV0 -0

+R1

kV0 -

k2

V0

Or kR1

(k - 1)V0 =kR2

V0 +k2R1

(k - 1)V0

Therefore, R2

R1 =k

(k -1) 2

Also, I’ = I’1 + I’

2

Or R1

kN-2

V0 -kN-1

V0

=R2

kN-1

V0 -0+

R1 + R3

kN-1

V0 -0

After Solving,

R3

R2 =k - 1

k

2) Here, I1 =R2

V1 -0 =R2

kV0 -0

I1 =kR2

V0 =k(

k - 1k

)R3

V0 =k2R3

(k - 1)V0