Connection of Capacitors

Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

CHAPTER III

THE ELECTRIC CURRENT Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

After completing this chapter the students should know:

 The concept of electric charge, electric force and electric field.

 The importance of Ohm’s law.

 The Kirchhoff’s law.

 The relation between electrical energy and thermal energy.

 The variation of resistance with temperature.

 The different types of resistors’s connection.

 The relation between electromotive force and internal resistance.

 The behaviour of charging and discharging circuits. Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

Chapter III The Electric Current 3.1: The Electric Current

The electric current can be expressed as follow:

Q I  t or (3.1) d Q I  d t

Where, dq is the charge and equals to dq  nev Adt (3.2)

Where: n is the No. of electrons e is the electronic charge v is the electron velocity

A is the area dt is the time

Then from equations (3.1) and (3.2), we can get

n e v Adt I  d t (3.3) I  nev A

Note that:

The current density (J) is defined as follow: Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

I n e v A J   A A (3.4) J n e v

If an electric field of strength (E) is applied on a conductor, the electrons will accelerate. Therefore, the electrical conductivity (σ) can be expressed as follow:

J   (3.5) E

Also, the specific resistivity of the material (ρ) is defined as follow

1   (3.6) 

E   J (3.7) E   J

But, as known:

V I E  and J  (3.8) d A

V is the applied voltage d is the distance

Then by substituting from equation (3.8) in equation (3.7), we can obtain

V I   d A (3.9)  d V  I A

The previous equation (3.9) can be take another form as follow:

V  R I (Ohm’s law) (3.10)

Where Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

R is the resistance of the material.

3:2 The Variation of Resistance with Temperature:

The resistance of the material can be expressed as a function of temperature as the following equation:

R  R0 (1  T) (3.11)

Where: R (Ω ) R is the resistance of the material at temperature T

0 R0 is the temperature of the material at T= 0 C. )R (Ω 0  is the temperature coefficient of resistance.

T oC Example (3.1):

A metallic wire with length of 100 m and cross-sectional area of mm2 carries a current of 20 A. Calculate the following:

1- The electric field (E).

2- The potential difference (V)

3- The resistance of the wire (R )

o o 4- The resistance of the wire (R0 and R100) at 0 C and 100 C.

Knowing that, the resistivity if the wire (=1.72x10-8 Ω cm) at T= 20 0C and α= 0.00393 C-1

Solution:

L= 100 m , A = 1mm2= (10-3)2= 10-6 m2 , I = 20 A

1- The electric field (E) is Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

I E   J   A

20 E  1.72x108 x  0.344 V / m 106 2- The potential difference (V) is V  E L  0.344 x 100 V  34.4 volt

3- The resistance (R ) is

V 34.4 R   I 20 R  1.72 

o o 4- The resistance of the wire (R0 and R100) at 0 C and 100 C are

R  R0 (1   T)

1.72 R0 (1  0.00393x 20)

R0  1.59 

0 Also the resistance (R100) at T= 100 C is

R100  R0 (1  T)

R100 1.59 (1  0.00393x100)

R100 2.22 

Example (3.2) A wire has a resistance of 5.4 Ω at temperature of 20 0C and another resistance of 7 Ω at temperature of 100 0C. Calculate the following:

1- The temperature coefficient of resistance ().

2- The value of R0

3- The temperature at which the resistance will be 8.8 Ω.

Solution

Suppose that:

0 R1= 5.4 Ω at T1= 20 C

0 R2= 7 Ω at T2= 100 C Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

R1  R0 (1   T1) 1- R2  R0 (1   T2 )

5.4 R (1  20 ) 0 7 R0 (1  100 )

Then by dividing, we can obtain:

5.4 R (1  20 )  0 7 R0 (1  1000 )

5.4 x (1  100 )  7 x (1  20 ) 5.4 540  7  140

 ( 540  140)  7 5.4 1.6   0.004 0C 1 400

2- The value of R0 can be calculated as follow:

5.4 R0 (1  20 x 0.004 ) 5.4 R   5  0 (1 20 x0.004) Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

3.3: Connection of Resistors

3.3:1 Resistors in Series

As shown in the figure, there are some resistors are connected in series.

R R R Note that: 1 2 3

V V V The applied voltage (V) on the resistors can be 1 2 3 I I I 2 expressed as follows: 1 3

A 3 B V

Vtotal V 1 V2 V3           Vn (3.12)

But according to Ohm’s law, the voltage on each resistor can be expressed as follows:

V  I R , V1  I 1 R1 , V2  I 2 R2 , V3  I 3 R3 (3.13)

Then by substituting from equation (3.13) in equation (3.12), we can get:

I total Rtotal  I 1 R1  I 2 R2  I 3 R3       I n Rn (3.14)

Note that:

In case of series connection, the current values are the same on all resistors, i.e.,

I total  I 1  I 2  I 3  I n

Then, the equation (3.14) will become

Rtotal  R1  R2  R3       Rn (3.15)

Or in general form, the total resistance of the circuit will be in the following form Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

n Rtotal   Ri (3.16) i1

3.3:2 Resistors in Series

As shown in the figure, there are some resistors are connected in R1 V1 I1 parallel.

R V I Note that: 2 2 2

V The total current (Itotal)in this circuit can be expressed R V I3 3 3 3 as follows: V 3 A B Itotal I 1  I2  I3            In (3.17) V

But according to Ohm’s law, the voltage on each resistor

can be expressed as follows:

Vt V1 V2 V3 It  , I1  , I 2  , I3  (3.18) Rt R 1 R2 R3

Then by substituting from equation (3.18) in equation (3.17), we can get:

V V V V V t  1  2  3          n (3.19) Rt R 1 R2 R3 Rn

Note that:

In case of parallel connection, the voltage values are the same on all resistors, i.e.,

V total  V 1  V 2  V 3 V n

Then, the equation (3.19) will become Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

1 1 1 1 1            (3.20) Rtotal R1 R2 R3 Rn

Or in general form, the total resistance of the circuit will be in the following form

1 n 1   (3.21) Rtotal i1 Rn

R =9Ω I 1 2 Example: (3.3) V 3 R =6Ω I 2 2 In the following figure, calculate: V R = 3 Ω3 3 1- The total resistance of this circuit. V 3 2- The total resistance of this circuit. V= 18 Volt 3- The current on each resistor.

Solution:

1- Since, the resistors are connected in parallel, hence

1 1 1 1    Rtotal R1 R2 R3

1 1 1 1    Rtotal 9 6 3

1 2  3 6 11   Rtotal 18 18

18 R   total 11 Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

2- The total current can be calculated according to Ohm’s law as follow:

Vtotal Itotal  Rtotal

18 I   11 A total (18/11)

3- The current on each resistor can be estimated as follow:

V1 18 I1    2 A R1 9

V2 18 I 2    3 A R2 6

V3 18 I3    6 A R3 3

Example: (3.4) R3 =9Ω I2 R 6 = Ω I 1 3 I 1.3 Ω 1 R 3= Ω a b c 4 d V VI I R 3 = Ω 3 34 I V 2 2 V 32 V V V 3 V R = 1.5 Ω V 5 V 3 3 V 3 3 3 V 3 3 I 3 5 V 3 V 3 V= 18 Volt

In the following figure, calculate:

1- The equivalent resistance of the circuit.

2- The total current value in the circuit.

3- The potential difference on each resistance. Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

4- The current value on each resistance.

Solution:

1- The equivalent resistance of the circuit can be calculated as follow: a- Firstly, the equivalent resistance between the points (b&c) is

1 1 1 1 2 3 1      Rbc 6 3 6 6 2

Rab 2 b- Firstly, the equivalent resistance between the points (c&d) is

1 1 1 1 1 36 10      Rcd 9 3 1.5 9 9

Rcd 0.9  3.1Ω R =2Ω R =0.9Ω I R = 6 Ω b bc cd d a Ω c Ω Ω V I V 3 V V 3 3 V V V V 3 3 V 3 V 18 V v 18 3 3 3 3

Therefore the equivalent resistance is

Rt  Rab  Rbc  Rcd  3.1  2 0.9 6

2- The total current (I) in the circuit

V 18 I    3 A R 6

3- Since, the resistors (Rab, Rbc and Rcd) are connected in series, the resistors have the same current. Therefore:

Vab  I x Rab  3 x3.19.3 volt

Vbc  I x Rbc  3 x26 volt

Vcd  I x Rcd  3 x0.92.7 volt

4- The current on each resistor is

 The ab branch Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

The current in the ab branch is

Vab 9.3 Iab    3 A Rab 3.1

 The bc branch

Since, the resistors are connected in parallel in the part of bc, the voltage values on all resistors are the same. Therefore: the values of current can be estimated as follow

V 6 I  ab   1 A 1 6 6

V 6 I  ab   2 A 2 3 3

 The bc branch

Also, in the part of cd, the values of the current can be estimated as follow:

V 2.7 I  cd  0.3 A 4 9 9 V 2.7 I  cd  0.9 A 5 3 3 V 2.7 I  cd  1.8 A 5 1.5 1.5

Example (3.5):

What resistance must be placed in parallel with 3 Ω to obtain a combined resistance of 5Ω.

Solution:

As known in case of parallel

  e Rtotal R1 R2 Then

1 1 1   3 5 R2 1 1 1 5  3 2     R2 3 5 15 15 Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

15 R  7.5 2 2

3.5: The Energy and Power in D.C Circuits

When an electric current (I) passes through a conductor of resistance (R), the electrical energy will convert to thermal energy. The value of charge (q) in the conductor within certain time of (t) can be written as follow

q  I t (3.22) and the electrical energy (U) which will be changed to thermal energy is

U  V q  V I t

U  I V t (3.23)

Also, the power (P) can be expressed as follow

U IV t P   t t (3.24) P  IV

But according to Ohm’s law

V  I R

P  IV  I 2 R (3.25) V 2  R

The unit of energy (U) is (watt. Sec) and the unit of power (P) is (watt).

Note that: Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

The electrical energy (U) is directly proportional to the thermal energy (H), according to the following equation:

U  H (3.26) U  J H

Where J is called as Joule’s constant (J= 4.18 Joules/cal). Then, the previous equation (3.26) can be written in another form as follow:

I V t  J m S (2 1 ) (3.27)

Where: m is the material’s mass

S is the specific heat of material

(θ1& θ2) are different temperatures

Example (3.4):

An electric heater with resistance of 20 Ω is connected to a power supply. Calculate the electrical power (P) if the passing current through it is 5A. Calculate the electrical energy (U), quantity of heat (H) and the cost of this heat after 15 days, knowing that, 1kW.h= 9 Halalah.

Solution:

The electrical power (P) can be calculated as follow:

P= I2 R = 52 x20= 500 W = 0.5 kW

The electrical energy (U) can be calculated as follow:

U = P t = 500x 15x24x60x60 = 6.48x108 J

U = 0.5x15x24= 180 kW .h Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

The quantity of heat (H) can be calculated as follow:

H = U/J = 6.48x108 /4.18= 1.55x108 Cal

The cost of this heat is

The cost = 180x9= 1620 Halalah

3: 6 The Electromotive Force and Internal Resistance

Note that, R

In the following figure, the total power (P) is r expressed as 

Pt  Pr  PR (3.28)

2 2 Pt = I PR= I R , Pr= I r

I   I 2 R  I 2 r

  I R  I r (3.29)   I ( R  r)

Hence:

The total current (I) can be expressed as follow

 I  (3.30) ( R  r)

VR = I R

Vr = I r

Example: (3.5) R=4Ω r=4Ω

= 12 V Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

In the following figure, calculate

1- The total current (I).

2- The voltage at each resistance.

Solution:

1- The total current can be calculated as follow

 I  ( R  r) 12 I   2 A 6

2- The voltage on each resistor

VR  I R  2x4 8 volt

Vr  I r

 2x2 4 volt Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

Kirchhoff Laws:

Since not all circuits can be reduced to simple series-parallel combinations, we need new rules to work with when we have resistance networks of cross connections such as the illustration at right.

In such a circuit, a branch point is a point where three or more conductors are joined. A loop is any closed conducting path. In the diagram, a, d, e, and b are branch points but c and f are not. aceda, defbd, and hadefbgh are examples of loops. Now, concerning networked cross connections, there are only two relatively simple rules known as Kirchhoff's rules (developed by Gustav Robert Kirchhoff (1824-1887)).

 Point Rule: The algebraic sum of the currents toward any branch point is zero.

 I 0

 Loop Rule: The algebraic sum of the potential differences in any loop, including those associated with EMFs and those of resistive elements, must equal zero.

  IR  0

Here is a good set of guidelines when you are working with these laws:

1. Choose any closed loop in the network and designate a direction to traverse the loop when you apply the loop rule (If you have a diagram, draw it on the diagram). Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

2. Go around the loop in that direction, adding EMF's and potential differences. An EMF is positive if you go from (-) to (+) (which is in the direction of the field E in the source) and negative when you go from (+) to (-). A -IR term is counted negative if the resistor is traversed in the same direction as the assumed current, positive if backwards. 3. Equate the sum of found in the previous step to zero. 4. If you need to, choose another loop to obtain a different relationship. The number of loop equations you have will be one less than the number of loops you have. You will be setting up a system of equations in order to find the current in each loop.

5. The last equation should be a branch equation, where I1 + I2 = I3 (substitute the currents for whichever branch you choose). 6. Solve the systems of equations. If one of the currents you solve is negative, it simply means that the current goes in the direction opposite the way you assumed. Remember, switch the current for only that current

Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

3. 6: Charging and Discharging Processes in RC Circuits

3.6.1: Charging Process

 Vc

In the charging process, the voltage of the battery () equals to

 VR VC (3.30)

VR is the potential difference across the resistance

VC is the potential difference across the capacitance

dq V  I R  R R dt (3.31) q V  C C

From equation (3.31) in equation (3.30), we can obtain:

dq q   R  dt C dq  C  RC  q (3.32) dt dq  C q  RC dt

Suppose that

y   C q dy dq (3.33)  dt dt

dy y  RC (3.34) dt

By using the variable’s separation property, Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

dy 1  dt y RC and by integration, we can obtain:

y dy 1 t  dt o y RC 0 t ln y   K , y  C q RC 1 t ln C q  K RC 1

To obtain the value of K1, we must use the boundary conditions as follows:

at t 0 and q  0  K1 ln C

t ln C q ln C RC t (3.35) ln C q ln C  RC

 C  q  t ln       C  RC  t      C  q   RC    exp   C   t      C  q  C exp RC 

 t     q   C   C exp RC 

 t       q   C 1exp RC        t       q  q 1exp RC   0    

Where:

q is the value of charge at any time (t) on the capacitor

q0 is the maximum value of charge on the capacitor

is the maximum voltage on the battery Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

Also, the current during the charging process can be calculated by differentiation equation (3.35) as follows:

t dq   I   (exp RC ) dt R t   I  I (exp RC ) , where I  (3.36) 0 0 R

3.6.2: Discharging Process

 Vc V R

In the discharging process, the voltage of the battery () equals to

 VR VC (3.37)

= 0 during the discharging

VR is the potential difference across the resistance

VC is the potential difference across the capacitance Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

dq V  I R  R R dt (3.38) q V  C C

dq q 0 R  dt C dq q R   dt C dq 1   dt q RC q dq 1 t   dt o q RC 0 t ln q    K (3.39) RC 1

To calculate the value of K1, the boundary conditions must be used as follows:

at t  0 and q q0  K1  ln q0

t ln q  ln q RC 0 t  q  t   ln q ln q0    ln  RC  q0  RC  t      RC  q q0 e (3.40)

The current during the discharging process can be estimated by the differentiation as follows:

 t  dq d      I   q e RC    0  dt dt    t  q    q  I  0 e RC  , 0   I RC RC R 0  t      RC  I I0 e (3.41) Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

In Summary, some equations about the charging and discharging processes.

Charging Process Discharging Process

 t  The Charge (q)       t  RC    q  q 1e    RC  0   q  q e   0

 t   t  The voltage (V)         v   1e RC    RC    v   e  

 t   t  The current (I)        RC  I   I e RC  I  I0 e 0

Also, note that :

In all the circuits of charging and discharging processes:

The Maximum Charge (q0) q0   C [C]

The Maximum Current (I )  0 I  [A] 0 R

The Time Constant (T) T  R C [sec]

Example (3.6): R=0.8 MΩ In the following figure, calculate:

1- The time constant (T). ε=12 V C= 5 F

2- The maximum value of charge (qo).

3- The maximum value of current (Io).

4- The charge (q) and current (I) as a function of time. Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

Solution:

1- The time constant (T) is

T= RC = 0.8x106x5x10-6 = 4 sec.

2- The maximum value of charge (qo) is

-6 -6 qo= ε C = 12 x5x10 = 60 x10 = 60 C

3- The maximum value of charge (Io) is

6 -6 Io= ε/R= 12/(0.8x10 ) = 15x10 A = 15 A.

4- The charge (q) and current (I) as a function of time are

 t RC q q0 (1 e ) t 60 (1 e 4 )  C and  t 4 I  I0 e  t 15 e 4  A

Example (3.7): R=5x103Ω Ω A) Calculate the required time to become the voltage across the capacitor is 99% from the ε=100 V R/=2x103 Ω Ω C= 20 F voltage of power supply. F

B) Calculate the current after 0.15 second, if the discharging process will taken place through the resistance (R/- 2x103 Ω).

Solution:

A) The voltage across the capacitor during the charging process is given by: Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

t     RC  V  1e   

t V   1e RC  t  99 3 6  1e 5x10 x20x10 100

t  99 e RC  1 100 t  10099 e RC  100

t  1 t e RC    ln1  ln 100 100 RC

t    2  t  2RC  2x5x103 x20x106  0.2 sec RC

B) The current can be calculated as follow:

t   I  e RC R 0.15  100 3 6 I  e 2x10 x 20 2x103

I 1.18x103 A

Example (3.8): R=100 MΩ

In the following figure, calculate the required ε C= 0.2 F time for the capacitor o save an electrical energy equals to a half of its maximum electrical energy. Electromagnetic 1 Chapter III Prof. Dr. T. Fahmy

Solution:

The electrical energy (U) can be expressed as follow:

1 U  q2 2C

But the charge (q) during the charging process is

t    q q 1 e RC  0    

Therefore:

t 2 1    U  q2 1 e RC  0   2C  

t 2     U 1 e RC  0    

t 2 1 1    U  U U  U 1 e RC  If 0 → 0 0   2 2  

Hence, t= 1.228

→→

RC = 1.228x100x106x0.2x10-6= 24.56 sec