CURRENT ELECTRICITY - Form 3 Physics Notes

• How to Use an Ammeter and Voltmeter
• Ohm's Law 
  • Factors Affecting the Resistance of a Conductor
• Resistors
  • Measurement of Resistance
  • Resistor Networks
• Internal resistance, r

How to Use an Ammeter and Voltmeter

• Connect the positive terminal of the ammeter/ voltmeter to the positive terminal of the battery.
• Ensure that the pointer is initially at zero i.e. there is no zero error. If there is a zero error, correct it before using the instrument.
• Select an appropriate scale to use.
• Avoid parallax error taking readings i.e. view the scale normally.

Ohm’s Law

• This law relates the current flowing through a conductor and the voltage drop across that section of the conductor.
• The law states: the current flowing through a conductor is directly proportional to the potential difference across its ends provided temperature and other physical factors are kept constant .
• The following set up can be used to investigate Ohm’s law:
  
  
  • Close the switch and adjust the current flowing through the conductor T using the rheostat to the least possible value. Record the corresponding voltmeter reading.
  • Increase the current in steps recording the corresponding voltmeter readings. Record your values in the table below
    

    Current I (A)
    Voltage V (V)

  • Plot a graph of voltage against current. Hence determine the slope of the graph.
• A graph of voltage against current is a straight line through the origin. Hence voltage drop across the conductor is directly proportional to the current through it;
  
  V α I
  ^V/_I = constant
• The constant is known as resistance R of the conductor T under investigation.
  Thus, V/I= R
  Or V= IR.
• Hence the slope of a voltage—current graph is equal to the resistance R of the conductor T.
• Electrical resistance can be defined as the opposition offered by a conductor to the flow of electric current. It is measured using an ohmmeter.
• The SI Unit of electrical resistance is the ohm (Ω). Other units include kilo-ohm (kΩ) and mega-ohm (MΩ);
  1Ω= 10^-3 kΩ
  1Ω= 10^-6 MΩ
• Materials which obey Ohm’s law are said to be ohmic materials while those which do not obey the law are said to be non-ohmic materials.
• The graph of voltage against current for non-ohmic materials is a curve or may be a straight line but does not pass through the origin.
• The inverse on resistance is called conductance;
  Conductance= ¹/_{resistance R.}

Example 5.3

1. Calculate the current flowing through a 8Ω device when it is connected to a 12V supply.
   I= ^V/_R
   I= ^12V/_8Ω =1.5A

Factors Affecting the Resistance of a Conductor

• There are three main factors that affect the resistance of a conductor:
  1. Temperature
     Increase in temperature enhances the vibration of the atoms and thus higher resistance to the flow of current.
  2. Length of the conductor L
     The resistance of a uniform conductor increases with increase in length.
  3. Cross section area A
     A conductor having a wider cross section area has more free electrons per unit length compared to a thin one.
     Hence a thicker material has a better conductivity than a thinner one. Generally, resistance varies inversely as the cross section area of the material.
• Therefore, at a constant temperature resistance varies directly as the length and inversely as the cross section area of the conductor;
  R α ^L/_A
  R= (A constant × ^L/_A)
  Or simply, ^AR/_L= constant
  The constant is called the resistivity of the material;
  Resistivity ϱ= ^{(cross section area A × resistance R)}/_{length L}.
  Resistivity is measured in ohm-metre (Ωm).

Example 5.4

1. A wire of resistance 3.5Ω has a length of 0.5m and cross section area 6.2 × 2 -8 m 5. Determine its resistivity.
   Resistivity ϱ= ^AR/_L = ^{(6.2 × 10-8m2 × 3.5Ω)}/_0.5m
   = 5.74 × 10^-7 Ωm
2. Two conductors A and B are such that the cross section area of A is twice that of B and the length of B is twice that of A. If the two are made from the same material, determine the ratio of the resistance of A to that of B.
   R=^ϱL/_A
   Therefore, RA = ^ϱ_AL_A/A_A
   And R_B = ^ϱ_BL_B/A_B
   Where L_B =2L_A
   A_B = ¹/2A_A
   And ϱ_A = ϱ_B
   Hence R_A =ϱ_AL_A/A_A and R_B = ^2ϱ_AL_A/0.5A_A = ^4ϱ_AL_A/A_A
   Thus ^R_A /R_B =^ϱ_AL_A/A_A = ¹/₄
   ^4ϱAL_A /A_A
   R_A : R_B = 1 : 4

Resistors

• A resistor is a specially designed conductor that offers a particular resistance to the flow of electric current.
• There are three main groups of resistors:
  1. Fixed resistors- offer fixed values of resistance. They have colour bands around them.
  2. Variable resistors- offer varying resistance e.g rheostat and potentiometer.
  3. Non-linear resistors- the current flowing through these resistors does not change linearly with the voltage applied. Examples include a thermistor and light-dependent resistor (LDR).

Measurement of resistance

• Three methods may be used:
  1. Voltmeter- ammeter method
     • In this method, the current flowing through the material and voltage across its ends are measured and a graph of voltage against current plotted.
     • The slope of the graph gives the resistance offered by the material.
  2. The wheatstone bridge method
     
     • A wheatstone bridge consists of four resistors and a galvanometer connected as shown below:
       
     • The values of three out of the four resistors must be known.
     • The value of one of the resistors is adjusted to a point that the galvanometer does not deflect. At this point, the voltage drop across R₁ is equal to that across R₃
     • Similarly, the voltage drop across R₂ is equal to that across R₁.
     • Note that the current flowing through R₂ is equal that through R₄. Also, the current through R₃ is the same to that through R₁.
       Therefore, I₁R₁ = I₂R₃ …………………………. (i)
       I₁R₂ = I₂R₄ …………………………. (ii)
       Dividing equation (i) by (ii), we get;
       ^R₁/R₂ = ^R₃/R₄
     • This method is more accurate compared to the voltmeter - ammeter method since the voltmeter has some resistance against the flow of current and thus takes up some voltage.
  3. The metre bridge method
     • This method relies on the fact that resistance is directly proportional to the length of the conductor.
       
     • The values of R₁ and R₂ must be known.
     • Suppose at point K the galvanometer does not deflect, then the voltage drop across R₁ equal the voltage drop across the section L₁ .
     • Similarly, the voltage drop across R 2 equals the voltage drop across the section L 5.
     • If the current through R 1 and R 2 is I 1 and that through the section L₁ and L₂ is I₂ , then;
       I₁R₁ = I₂L₁ ………………………….. (i)
       I₁R 2 = I₂L₂ …………………………… (ii)
       Dividing equation (i) by (ii), we get;
       ^R₁/R₂ = ^L₁/L₂

Example 5.5

1. In an experiment to determine the resistance of a nichrome wire using the metre bridge, the balance point was found to be at the 40cm mark. Given that the value of the resistor to the right is 30Ω, calculate the value of the unknown resistor R.
   
   ^L_AC /L_CB = ^R/_30Ω
   ^40cm/_60cm = ^R/_30Ω
   R= ^(30×40)/₆₀ = 20Ω

Resistor networks

Series network

• When resistors are arranged in series the same current pass through each one of them.
• Consider three resistors connected as shown below:
  
  From Ohm’s law, V= IR.
  The voltage drop across R₁; V₁ = IR₁
  The voltage drop across R₂; V₂ = IR₂
  The voltage drop across R₃; V₃ = IR₃
  And the total circuit voltage V= V₁ +V₂ +V₃ .
  Thus V = IR₁ +IR₂ +IR₃ =I(R₁ +R₂ +R₃)
  ^V/_I =(R₁ + R₂ + R₃ )
  But ^V/_I = R
  Thus the combined circuit resistance R = R₁ +R₂ + R₃ .
• Generally, the effective resistance of resistors arranged in series is equal to the sum of the individual resistances.

Parallel network

• When resistors are connected in parallel, the same voltage is dropped across them.
• Consider three resistors connected as shown below:
  
• Suppose the current flowing through R 1 is I₁ , through R₂ is I₂ and through R₃ is I₃ then:
  The voltage drop across R₁; V₁ = I₁R₁
  The voltage drop across R₂; V₂ = I₂R₂
  The voltage drop across R₃; V₃ =I₃R₃
  But V₁ = V₂ = V₃ = V and I = I₁ +I₂ +I₃
  Therefore, I=^V/R₁ + ^V/R₂ + ^V/R₃
  ^I/_V = (¹/R₁ + ¹/R₂ + ¹/R₃)
  But ^I/_V= ¹/_R.
  Hence ¹/_R= ¹/R₁ + ¹/R₂ + ¹/R₃
  R is the combined circuit resistance.
• Special case of two resistors in parallel
  It follows that ¹/_R= ¹/R₁ +¹/R₂
  1/R= ^{(R₁ + R₂ )}/R₁R₂
  Hence the effective resistance R= ^R₁R₂/_{(R1 + R2)}.
• Generally for n resistors arranged in parallel, the effective resistance of the arrangement is given by;
  ¹/_R=¹/R₁ + ¹/R₂ +…………..+¹/R_n
• NOTE: when a circuit comprise of both series and parallel connections, the arrangement is systematically reduced to a single resistor.

Example 5.6

1. The figure below shows 3 resistors.
   
   Calculate:
   1. The effective resistance of the circuit.
      R= (8+5+3)Ω = 3Ω
   2. The total current in the circuit.
      I=^V/_R = ^12V/_3Ω = 0.75A
   3. The voltage drop across each resistor.
      V_8Ω = 0.75 × 8 = 5.0V
      V_5Ω = 0.75 × 5 = 3.75V
      V_3Ω =0.75 × 3 = 5.25V
2. Three resistors are connected as shown below:
   
   Calculate:
   1. The total resistance of the circuit.
      ¹/_R= ¹/₅ + ¹/₃ + ¹/₆
      ¹/_R= ^(6+2+5)/₃₀ =²¹/₃₀
      R= ³⁰/₂₁ = 1.4286Ω
   2. The current through each resistor.
      I_5Ω =^12V/_5Ω=5.4A
      I_3Ω =^12V/_3Ω=1.0A
      I_6Ω =^12V/_6Ω=5.0A
3. The figure below shows five resistors and 6.0V supply.
   
   Calculate:
   1. The effective resistance of the circuit.
      R_2,3Ω =^(2×3)/₍₂₊₃₎ = 1.2Ω
      R_4,1.2,0.2Ω =4 + 1.2+0.2 =5.4Ω
      R= R_1,5.4Ω =^(1×5.4)/_(1+5.4) = 0.8438Ω
   2. The total circuit current.
      I=^V/_R = ^6V/_0.8438Ω =3.127A

Internal Resistance r

• When a cell supplies current in a circuit, the potential difference between its terminals is observed to be lower than its electromotive force (emf).
• This difference is due to the internal resistance of the cell.
• Some work must be done to overcome this resistance and so the drop in the emf of the cell is responsible for this.
• The difference is referred to as the lost volt and is given by Ir.
  i.e. lost volts = emf − terminal voltage
  Or simply emf = terminal voltage + lost volts
• The mathematical equation connecting emf, circuit current, external resistance and internal resistance of the cell is given by:
  E= IR + Ir= I(R+r).
• Internal resistance of a cell can be obtained experimentally. In such an experiment, the following data was obtained:
  

  Current I (A)	0.1	0.2	0.3	0.4	0.6	0.8
  Volatge V (V)	1.43	1.30	1.4	1.09	0.82	0.58

• When a graph of Voltage V against current I is plotted, the graph will appear as shown below: