Unit II: Current Electricity · Current Electricity UnitII:CurrentElectricity Dr Sukanta Deb...

Current Electricity

Unit II: Current Electricity

Dr Sukanta Deb

Department of Physics, Cotton UniversityPanbazar, Guwahati (Assam)

Subject: Physics, HS 2nd Year

Module: II

Dr Sukanta Deb Unit II: Current Electricity 1 / 45

Current Electricity

Outline

1 Current Electricity


Current Electricity

Learning Objectives (Module: II)

1 Carbon ResistorsColour Code for Carbon ResistorsTemperature Dependence of ResistanceSeries and Parallel Combinations of Resistors

2 Internal Resistance of a CellPotential Difference and EMF of a CellCombination of Cells in Series and Parallel


Current Electricity

Carbon ResistorCarbon resistor consists of a ceramiccore (insulator), on which a thin layer ofcrystalline carbon (conductor) isdeposited. The ratio of carbon toceramic (conductor to insulator)determines the overall resistive value ofthe mixture. Higher the ratio of carbon,the lower the overall resistance.The mixture is moulded into acylindrical shape with metal wires orleads attached to each end to providethe electrical connection. Then it iscoated with an outer insulating materialsuch as ceramic or plastic and colourcoded with bands to denote its resistivevalue in ohm.

Figure: Carbon resistors usedin our laboratory.

The resistive valuesassociated with each ofthe colour codes are shownin the following table.


Current Electricity

Carbon Resistor: Colour Codes

Each colour is given a specificnumber, a multiplier and atolerance associated with it.Tolerance of a resistor is thepercentage error in the resistorvalue.It is always convenient toremember the sequence“BBROYGBVGW” in order ofascending colour code numbers 0to 9 assigned to the respectivecolours. One of the very widelyused mnemonics to remember thesequence is “BB ROY of GreatBritain has a Very Good Wife”. Figure: Colour coding for resistors.


Current Electricity

Colour Coding for ResistorsThe value of the resistance of thecarbon resistor is indicated by thefour coloured bands, marked onthe surface of the cylinder. Themeanings of the four positions ofthe bands are shown in Figure.The first three coloured bandsare used to indicate the values ofa resistor:

+ the first two bands aresignificant digits ofresistances,

+ the third band indicates thedecimal multiplier afterthem.

Figure: Colour coding for resistors.

The fourth colour, silver or gold,shows the tolerance of theresistor at 10% or 5% as shownin the Figure. If there is nofourth band, the tolerance is20%.


Current Electricity

Finding Resistance Using Colour Code

Steps1 Always place the resistor in front of you such that the colour

bands lie to your left. In this position start reading theresistance values. The last band indicates the tolerance valueof the resistor.

2 Start writing the value of colour codes from the left band withthe last band before the tolerance band as multiplier.

For the resistor as shown in the earlier figure, the first digit = 5(green), the second digit = 6 (blue), decimal multiplier=103

(orange), Tolerance=5%(gold). The value of the resistance is

R = (digit 1 digit 2)× decimal multiplier± tolerance⇒ R = (56)× 103 ± 5% = (56000± 5%) Ω⇒ R = (56000± 2800) = (56.0± 2.8) kΩ ∵ 5% of 56000 = 2800.


Current Electricity

Problem Based on Colour Code

1 How will you represent a resistance of 3700Ω± 10% by colourcode?Given

R =3700Ω± 10%

⇒ R = 3︸︷︷︸orange

7︸︷︷︸violet

×102︸︷︷︸

red ± 10%︸︷︷︸silver

.

The colours attached to numbers 3 (digit 1), 7 (digit 2) and2 (decimal multiplier) are orange, violet and red, respectively.A tolerance value of 10% represents silver. Thus the bands ofcolour on the resistor in sequence are orange, violet, red andsilver.


Current Electricity

Effect of Temperature on ResistanceWe know that the resistance of a metallic conductor is given by

R =ρ lA

⇒ R = 1σ

lA ∵ ρ = 1

σ

⇒ R = 1(ne2τ

m

) lA ∵ σ = ne2τ

m

⇒ R =1τ

( mne2

lA

).

For a given conductor the quantity(

mne2

lA

)=constant. Therefore

we have

R ∝ 1τ.


Current Electricity

Effect of Temperature on Resistance (contd...)When the temperature of a conductor is increased the thermalenergy of the conductor increases and as a result the collisionfrequencies of free electrons with the positive ions increase. Thisdecreases the mean free time (τ) and hence the resistance of theconductor increases as R ∝ 1

τ .Let the change in resistance be ∆R for a small change intemperature ∆T . The fractional change in resistance (∆R

R ) isdirectly proportional to the change in temperature (∆T ), .i.e.,

∆RR ∝∆T

⇒ ∆RR =α∆T ,

where α is called the temperature coefficient of resistance and ischaracteristic of the material of the conductor. Its value variesfrom metal to metal and in general depends on temperature.


Current Electricity

Effect of Temperature on Resistance (contd...)∆R is the change in resistance, R is the original resistance, ∆T isthe change in temperature.The above equation can be written as

α = ∆RR∆T .

Thus the temperature coefficient of resistance (α) is defined as theincrease in resistance per unit original resistance per degree riserise in temperature. The unit of α is C−1 or K−1.Let R1 and R2 be the resistances at temperatures T1 and T2,respectively (where T1 < T2), Then ∆R = R2 − R1 and∆T = T2 − T1 and R = R1. Therefore we have

∆RR =α∆T

⇒ R2 − R1R1

=α (T2 − T1)

⇒ R2 − R1 =R1α (T2 − T1)Dr Sukanta Deb Unit II: Current Electricity 12 / 45

Current Electricity

Effect of Temperature on Resistance (contd...)

⇒ R2 =R1 + R1α (T2 − T1)⇒ R2 =R1 [1 + α (T2 − T1)] . (1)

If we write T1 = T0 (any reference temperature) and R1 = R(T0)and T2 = T , R2 = R(T ), then the equation (1) can be written as

R(T ) =R(T0) [1 + α (T − T0)] (2)⇒ R(T ) = R(T0) [1 + α∆T ] , (3)

where ∆T = T − T0. We know that

R = ρlA .

Using this formula in equation (3), we get

ρ(T )lA =ρ(T0)

lA (1 + α∆T )

⇒ ρ(T ) =ρ(T0) (1 + α∆T )⇒ ρ(T ) =ρ(T0) [1 + α∆T ] .Dr Sukanta Deb Unit II: Current Electricity 13 / 45

Current Electricity

Effect of Temperature on Resistance (contd...)If T0 = 0C, then we haveR(T ) = R0 (1 + αT ) and ρ(T ) = ρ0 (1 + αT ), where R0 and ρ0denote the resistance and resistivity, respectively at 0C.

1 In metals resistivity increases with increasing temperature andhence α is positive. If the temperature of a conductorincreases, the average kinetic energy of electrons in theconductor increases. This results in more frequent collisionsand hence the resistivity increases.

2 In semiconductors, resistivity decreases with increasingtemperature and hence α is negative. As the temperatureincreases, more electrons will be liberated from their atoms forconduction in semi conductors. Hence the current increasesand therefore the resistivity decreases. A semiconductor witha negative temperature coefficient of resistance is called athermistor.


Current Electricity

Problems Based on Effect of Temperature on Resistance

Formula Used+ For resistance:

1 R(T ) = R0 (1 + αT ), R0 is the resistance at 0 C.2 R2 = R1 [1 + α (T2 − T1)]3 α = R2−R1

R1(T2−T1) .+ For resistivity:

1 ρ(T ) = ρ0 (1 + αT ), ρ0 is the resistivity at 0 C.2 ρ2 = ρ1 [1 + α (T2 − T1)]3 α = ρ2−ρ1

ρ1(T2−T1) .


Current Electricity


1 The resistance of a conductor at 20C is 3Ω and at 100C is6Ω. Determine the temperature coefficient of resistance of theconductor. What will be the resistance of the conductor at0 C?Given T1 = 20 C, T2 = 100 C and R(T1) = R(20) = 3Ω,R(T2) = R(100) = 6Ω. Here we will apply the formulaR(T ) = R0 (1 + αT ), where R0 is the resistance at 0C andR(T ) is the temperature at T C. We have

R(T1) = R(20) =R0 (1 + 20α)R(T2) = R(100) =R0 (1 + 100α) .

ThereforeR(T2)R(T1) =R(100)

R(20) = R0 (1 + 20α)R0 (1 + 100α)


Current Electricity


⇒ 63 =(1 + 100α)

(1 + 20α)

⇒ 2 =(1 + 100α)(1 + 20α) ⇒ 1 + 100α = 2 (1 + 20α)⇒ 60α = 1

⇒ α = 160 .

Therefore we have

R(T1) =R0 (1 + 20α)⇒ 3 = R0

(1 + 20 1

60

)⇒ R0

(1 + 1

3

)=3⇒ R0

(43

)= 3⇒ R0 = 9

4 = 2.5Ω.


Current Electricity


2 The resistance of a heating element is 100 Ω at 20C. What isthe temperature of the element if its resistance is 117 Ω.Temperature coefficient of resistance = 1.7× 10−4 C−1.Given T1 = 27C, R1 = 100Ω and R2 = 117Ω,α = 1.7× 10−4 C−1, T2=? We know that

R2 =R1 [1 + α (T2 − T1)]

⇒ T2 − T1 =R2 − R1αR1

⇒ T2 − T1 = 117− 100(1.7× 10−4)× 100 = 17

(1.7× 10−4)× 100⇒ T2 − T1 =1000

⇒ T2 = 1000 + T1 =1000 + 27 = 1027C.


Current Electricity


3 The resistance of a wire is 3.00 Ω at 0C and 3.75 Ω at100C. Its resistance is measured to be 3.15Ω at roomtemperature. Find the room temperature.We know that

α = R2 − R1R1 (T2 − T1) = 3.75− 3.00

3.00 (100− 0) = 0.0025C−1.

We also know that

R(T ) =R0 (1 + αT )

⇒ T =R(T )− R0αR0

= 3.15− 3.000.0025× 3.00 = 20C.


Current Electricity

Series and Parallel Combination of Resistors

1 Resistances in Series:When resistors are connected in series(end to end), the equivalent resistance ofthe combination is equal to the sum ofthe individual resistances, i.e.

Req =R1 + R2 + R3 + . . .

The current is the same in all resistors.The total potential difference across thecombination is equal to the sum of thepotential differences across the individualresistors, i.e.

V =V1 + V2 + V3 + . . .

Figure: Resistances in series.

Note: When resistors areconnected in series, theequivalent resistance in thecircuit will be greater thaneach individual resistance.


Current Electricity

Series and Parallel Combination of Resistors

2 Resistances in Parallel:When resistors are connected in parallel,the equivalent resistance of thecombination is given by

1Req

= 1R1

+ 1R2

+ 1R3

+ . . .

The potential difference is same acrosseach resistor. The total current is equalto the sum of the currents in theindividual resistors, i.e.

I =I1 + I2 + I3 + . . .

Figure: Resistances inparallel.

Note: When resistors areconnected in parallel, theequivalent resistance in thecircuit will be lesser thaneach individual resistance.


Current Electricity

Series Combination of ResistorsThe Figure shows three resistors havingresistances R1,R2 and R3 connected in series.A potential difference V is applied betweenthe two end points of the combination. Thesame current I passes through all the resistors.Let V1,V2 and V3 be the potential difference(voltage) across each of the resistor.According to the Ohm’s law

V1 =IR1, V2 = IR2 and V3 = IR3.

The potential difference across thecombination is equal to the sum of thepotential differences across each of theresistors, i.e,

V =V1 + V2 + V3

Figure: Resistances in series.

⇒ V =IR1 + IR2 + IR3

⇒ V =I (R1 + R2 + R3)

⇒ VI =R1 + R2 + R3

⇒ Req =R1 + R2 + R3.


Current Electricity

Series Combination of ResistorsFor any number of resistors connected in series, the equivalentresistance of the combination is given by

Req =R1 + R2 + R3 + . . .

Thus the equivalent resistance of a number of resistors connectedin series is equal to the sum of individual resistances.


Current Electricity

Parallel Combination of Resistors

The Figure shows three resistors havingresistances R1,R2 and R3 connected inparallel. The potential difference V acrosseach of the resistors is the same. The totalcurrent I that leaves the battery is split intothree separate paths. Let I1, I2 and I3 be thecurrent through the resistors R1,R2 and R3,respectively. Therefore

I =I1 + I2 + I3.

Since the voltage across each resistor is thesame, applying the Ohm’s law we have

I1 = VR1, I2 = V

R2, I3 = V

R3.

Figure: Resistances in parallel.

Therefore,

⇒ I =I1 + I2 + I3

⇒ I = VR1

+ VR2

+ VR3


Current Electricity

Parallel Combination of Resistors

⇒ IV = 1

R1+ 1

R2+ 1

R3

⇒ 1Req

= 1R1

+ 1R2

+ 1R3.

For any number of resistors connected in parallel, the equivalentresistance of the combination is given by

1Req

= 1R1

+ 1R2

+ 1R3

+ . . .

Thus the reciprocal of equivalent resistance of a number ofresistors connected in parallel is equal to the sum of the reciprocalsof the individual resistances.


Current Electricity

Problems Based on Combinations of Resistors

1 Two electric bulbs marked 20 W-220 V and 100 W-220 V areconnected in series to 440 V supply. Which bulb will be fused?To check which bulb will be fused, the voltage drop acrosseach bulb has to be calculated. The resistance of a bulb isgiven by

R =V 2

P = (Rated voltage)2

Rated power .

Let 1 and 2 denote the first and the second bulb, respectively.Given P1 = 20 W, V1 = 220 V; P2 = 100 W , V2 = 220 V.Supply voltage V = 440 V. Therefore we have

R1 =V 21

P1= 2202

20 = 2420 Ω and R2 = V 22

P2= 2202

100 = 484 Ω.

Since both the bulbs are connected in series, the currentpassing through them will be the same.


Current Electricity


The current that passes through the circuit is given by

I = VReq

,

whereReq =R1 + R2 = (2420 + 484) Ω = 2904 Ω.

Therefore

I = 4402904 = 0.151 A.

The voltage drop across the first bulb is

Vdrop,1 =IR1 = 4402904 × 2420 = 366.7 V.

The voltage drop across the second bulb is

Vdrop,2 =IR2 = 4402904 × 484 = 73.3 V.


Current Electricity


Now we compare the voltage rating of each bulb with thecorresponding voltage drop across it. We see thatV1 < Vdrop,1 = 220 < 366.7 and V2 > Vdrop,2 = 220 > 73.3.Since V1 < Vdrop,1, the bulb 1 (P1 = 20 W,V1 = 220 V) willbe fused.

2 Calculate the equivalent resistance for the circuit which isconnected to 24 V battery and also find the potentialdifference across 4Ω and 6Ω resistors in the circuit.


Current Electricity


Given R1 = 4Ω, R2 = 6Ω and V = 24 V. Let V1 and V2denote the potential differences across the two resistors withresistances R1 and R2, respectively. We have to find thevalues of V1 and V2.Since the resistors are connected in series, the equivalentresistance of the combination is

Req =R1 + R2 = (4 + 6) Ω = 10Ω.

The current in the circuit is given by

I = VReq

= 2410 = 2.4 A.

Therefore we have

V1 = IR1 = 2.4× 4 = 9.6 V and V2 = IR2 = 2.4× 6 = 14.4 V.


Current Electricity


3 Calculate the equivalent resistancein the following circuit and alsofind the current I , I1 and I2 in thegiven circuit.

Since the resistances are connected inparallel, the equivalent resistance of thecombination is given by

1Req

= 1R1

+ 1R2⇒ Req = R1R2

R1 + R2

⇒ Req =4× 64 + 6 = 24

10 = 2.4 Ω.

Since the resistances are connected inparallel, the potential (voltage) acrosseach resistor is the same.

I1 = VR1

= 244 = 6 A,

I2 = VR2

= 246 = 4 A

∴ I =I1 + I2 = 6 + 4 = 10 A.


Current Electricity

Internal Resistance of a cell

An electric cell converts chemicalenergy into electrical energy to produceelectricity. It contains two electrodesimmersed in an electrolyte as shown inthe Figure. Internal resistance of a cellis defined as the resistance offered bythe electrolyte and electrodes of a cellwhen the electric current flows throughit. Internal resistance of a cell dependsupon the following factors:

1 Distance between the electrodes2 The nature of the electrolyte3 The nature of electrodes

Figure: A simple electric cell.

4 Area of electrodes immersedin electrolyte. if areaincreases, internal resistancedecreases.


Current Electricity

Potential Difference and EMF of a Cell1 A cell is a source of electromotive force (emf). The potential

difference between the terminals (the two electrodes) of a cellwhen it is on an open circuit, i.e. the terminals are notconnected externally is called its emf. It is denoted by E andis nonelectrostatic in nature.

2 The term ‘electromotive force’ is a misnomer since it does notreally refer to a force but describes a potential difference(work done/charge) in volts. The emf determines the amountof work done a cell does to move a certain amount of chargearound the circuit.

3 The potential difference between the terminals of a cell whenit is in a closed circuit, i.e. the terminals are connectedexternally by a conducting wire is called its terminal voltage(V ).

Note:Several electric cells connected together form a battery.Dr Sukanta Deb Unit II: Current Electricity 32 / 45

Current Electricity

Potential Difference and EMF of a Cell

Let us consider a cell having resistance freeemf E in series with an internal resistance r asshown by the dashed rectangle. An externalresistance (often called the load resistance) Ris connected to the terminals of the cell.A positive charge while passing from thenegative terminal (at a) to the positiveterminal (at b) gains potential by an amountE . However while going through the resistancer (from point c to d), its potential decreasesby an amount Ir , where I is the current in thecircuit. The terminal potential difference Vadis defined as the potential difference betweenthe terminals a and d of the cell, i.e.

Figure: Circuit diagram of acell of internal resistance r ,connected to an externalresistor of resistance R.

Vad = E − Ir . (4)

When I = 0 (i.e., open-circuited), Vad = E .


Current Electricity

Potential Difference and EMF of a CellThus we find that the emf of a cell is equal to the potentialdifference between the terminals when the terminals areopen-circuited. When the load resistance R is connected to theterminals, the potential difference across R is the terminalvoltage,i.e.,

Vad =IR.Therefore equation (4) becomes

IR =E − Ir

⇒ I (R + r) =E ⇒ I = ER + r .

Therefore the terminal potential difference is given by

Vad =( E

R + r

)R

⇒ Vad =( R

R + r

)E .


Current Electricity

Potential Difference and EMF of a Cell

Total Power Output Associated with the emfWe know that

P =IE⇒ P =I × (IR + Ir)⇒ P =I 2R + I 2r .

Here I 2R is the power delivered to the external load resistance Rand I 2r is the power delivered to the internal resistance r . For agood cell, r << R and hence I 2r << I 2R and almost entirepower is delivered to the load resistance R.

1 A cell has an emf of 12.0 V and an internal resistance of0.05Ω. Its terminals are connected to a load resistance of 3Ω.(a) Find the current in the circuit and the terminal voltage ofthe cell.


Current Electricity

Problems Based on EMF

(b) Calculate the power delivered to the load resistor, the powerdelivered to the internal resistance of the battery, and the powerdelivered by the cell.(a) The current in the circuit: I = E

R+r = 12.03.00+0.05 = 3.93 A.

The terminal voltage: V = E − Ir = 12.0− (3.93) (0.05) = 11.8 V.To check this result, calculate the voltage across the loadresistance R: VR = IR = 3.93× 3.00 = 11.8 V.(b) Power delivered to the load resistor:PR = I 2R = 3.932 × 3.00 = 46.3 W.Power delivered to the internal resistance:Pr = I 2r = 3.932 × 0.05 = 0.772 W.Power delivered by the cell:P = PR + Pr = 46.3 + 0.772 = 47.1 W


Current Electricity

Combination of Cells in SeriesCells are said to be connected in serieswhen they are joined end to end so thatthe same current flows through eachcell.In series connection, the negativeterminal of one cell is connected to thepositive terminal of the second cell, thenegative terminal of second cell isconnected to the positive terminal ofthe third cell and so on. The freepositive terminal of the first cell and thefree negative terminal of the last cellbecome the terminals of the battery.Let us consider three cells with emfsE1, E2 and E3 having internal resistances

Figure: Three cells in series.

r1, r2 and r3 connected inseries. Suppose anexternal resistance R isconnected across thecombination.


Current Electricity

Combination of Cells in SeriesThe free positive terminal of the first cell and the free negativeterminal of the third cell act as the terminals of the combination.Let I be the current flowing through the cells. Let V1,V2 and V3denote the terminal potential differences across the three cells,respectively. Then the potential difference (V ) between the twoterminals of the combination is given by

V =V1 + V2 + V3

⇒ IR = (E1 − Ir1) + (E2 − Ir2) + (E3 − Ir3)⇒ IR = (E1 + E2 + E3)− I (r1 + r2 + r3)

⇒ I [R + (r1 + r2 + r3)] =E1 + E2 + E3

⇒ I = E1 + E2 + E3R + (r1 + r2 + r3) = E

R + r .

Thus we see that the combination acts as a battery of emfE = E1 + E2 + E3 having an internal resistance r = r1 + r2 + r3.


Current Electricity

Combination of Cells in ParallelCells are said to be connected in parallelwhen the current is divided among thevarious cells.In parallel connection all the positiveterminals of the cells are connected toone point and all the negative terminalsto a second point. These two pointsform the positive and negative terminalsof the combination. Let us considerthree cells with emfs E1, E2 and E3having internal resistances r1, r2 and r3connected in parallel. Suppose anexternal resistance R is connectedacross the combination. Let I be thetotal current flowing in the externalcircuit.

Figure: Three cells in parallel.

Let this current be dividedinto three currents I1, I2 andI3 across the three cells,respectively.


Current Electricity

Combination of Cells in ParallelThe terminal voltage is the same across all the three cells. Let Vdenote the common terminal voltage. Therefore we have for thethree cells,

V =E1 − I1r1 = E2 − I2r2 = E3 − I3r3.

This gives

I1 =E1 −Vr1

, I2 = E2 −Vr2

, I3 = E3 −Vr3

.

Therefore the total current (I ) is given by

I =I1 + I2 + I3 =(E1 −V

r1

)+(E2 −V

r2

)+(E3 −V

r3

)⇒ I =

(E1r1

+ E2r2

+ E3r3

)−V

( 1r1

+ 1r2

+ 1r3

).


Current Electricity

Combination of Cells in Parallel

⇒ I =(E1

r1+ E2

r2+ E2

r2

)− IR

( 1r1

+ 1r2

+ 1r3

)⇒ I + IR

( 1r1

+ 1r2

+ 1r3

)=(E1

r1+ E2

r2+ E3

r3

)∵ V = IR

⇒ I[1 + R

( 1r1

+ 1r2

+ 1r3

)]=E1r2r3 + E2r1r3 + E3r1r2

r1r2r3

⇒ I[1 + R

(r2r3 + r1r3 + r1r2r1r2r3

)]=E1r2r3 + E2r1r3 + E3r1r2

r1r2r3

⇒ I[r1r2r3 + R (r2r3 + r1r3 + r1r2)

r1r2r3

]=E1r2r3 + E2r1r3 + E3r1r2

r1r2r3

⇒ I = E1r2r3 + E2r1r3 + E3r1r2r1r2r3 + R (r1r2 + r2r3 + r3r1)

⇒ I =E1r2r3+E2r1r3+E3r1r2

r1r2+r2r3+r3r1(r1r2r3

r1r2+r2r3+r3r1

)+ R


Current Electricity

Combination of Cells in Parallel

I = Er + R .

Thus we see that the combination acts as a cell of emf

E =E1r2r3 + E2r1r3 + E3r1r2r1r2 + r2r3 + r3r1

and internal resistance

r = r1r2r3r1r2 + r2r3 + r3r1

⇒ 1r =r1r2 + r2r3 + r3r1

r1r2r3

⇒ 1r = 1

r1+ 1

r2+ 1

r3.


Current Electricity

Combination of Two Cells in ParallelNote that if we have only two cells with emfs E1, E2 havinginternal resistances r1, r2 then we can write

I1 =E1 −Vr1

and I2 = E2 −Vr2

The total current (I ) is given by

I =I1 + I2 = E1 −Vr1

+ E2 −Vr2

⇒ I =(E1

r1+ E2

r2

)−V

( 1r1

+ 1r2

)⇒ I =

(E1r1

+ E2r2

)− IR

( 1r1

+ 1r2

)⇒ I + IR

( 1r1

+ 1r2

)=(E1

r1+ E2

r2

)⇒ I

[1 + R

( 1r1

+ 1r2

)]=E1r2 + E2r1

r1r2


Current Electricity

Combination of Two Cells in Parallel

I[1 + R (r2 + r1)

r1r2

]=E1r2 + E2r1

r1r2

⇒ I[r1r2 + R (r1 + r2)

r1r2

]=E1r2 + E2r1

r1r2

⇒ I [r1r2 + R (r1 + r2)] =E1r2 + E2r1 ⇒ I = E1r2 + E2r1r1r2 + R (r1 + r2)

⇒ I =E1r2+E2r1

r1+r2r1r2

r1+r2+ R ⇒ I = E

r + R .

Thus we see that the combination of two cells act as a cell of emf

E =E1r2 + E2r1r1 + r2

and internal resistance

r = r1r2r1 + r2

⇒ 1r = r1 + r2

r1r2⇒ 1

r = 1r1

+ 1r2.


Current Electricity

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