Potentiometer as a Voltage Divider

The rate of flow of electrons across a conductor is known as electric current. It can be simply defined as the amount of charge flown through a conductor upon time period. So, we can generate Formulae for Electric current from the above definition as

I = q/t

where,

q= charge

t= time

SI unit of electric current is Ampere(A).

1 Ampere is defined as 1 coulomb of charge passing through a conductor in 1 second.

The direction of the flow of electric current is opposite to that of the flow of electrons.

Types of Electric current

Electric current can be of two types depending on its magnitude and direction.

Direct Current

A type of current in which magnitude and direction remain the same with respect to time is known as a Direct current. In a Direct Current, a steady amount of current flows through a conductor, and there is no fluctuation in magnitude of current i.e. magnitude is constant, not variable. 

For eg. Cell, Battery, etc.

Alternating Current

A current whose magnitude is variable and direction reverse periodically is known as an Alternating current. In Alternating current, magnitude first increases from 0 to maximum gradually and then again decreases to 0. Then again increases to maximum but in the negative direction and then again comes to zero. This completes one cycle of alternating current.

The above process is so fast that there is any number of cycles in a single second.

The number of cycles per second is measured in Hertz (Hz). Like electricity supply in India comes at 50Hz. It means 50 cycles are completed in one second.

For eg. AC Dynamo, Supply to our homes through the powerhouse.

Electrical Resistance

The opposition to the flow of electric current across a conductor is known as electrical resistance. Ohm’s law states that potential difference across the conductor is directly proportional to the current flowing through the conductor.

V ∝ I 

V = IR

where R is the proportionality constant which is Resistance.

Formulae for electrical resistance can be concluded from ohm’s law as, R = V/I 

where V = Potential Difference 

I = Electric Current

The SI unit of resistance is Ohm.

1 ohm is defined as resistance offered by a conductor when a potential difference of 1 volt is applied across the conductor and the current flowing through the conductor is 1 Ampere.

Resistance can be represented as a Zig- zag line in a circuit. 

Potential Difference

The difference in potential between two points represents the measure of energy involved or the energy released in the transfer of the charge carriers from one point to the other. It is defined as the product of resistance across the wire and current flowing through the wire. From the above definition following formulae can be generated, V = IR

The SI unit of potential difference is Volt (V).

Potentiometer

A Potentiometer is a three-terminal resistor with two of its terminal fixed and one terminal variable or moving. if we break the word “potentiometer”, we will get two words “potential” and “meter”. So, it becomes pretty clear that a potentiometer is a device that is used to measure the potential difference between conductors.

Principle The principle behind a potentiometer is that it finds the value of an unknown voltage by comparing it with a known voltage.

Voltmeter and Potentiometer

We should not often confuse between voltmeter and potentiometer. They are different things. The voltmeter is used to find the terminal voltage of the circuit whereas the Potentiometer is used to find the emf of the circuit.

Applications

1. A potentiometer is used to find the unknown voltage by comparing it with the known voltage.
2. It is used to determine the value of emf and internal resistance of the given cell.
3. It can also be used as a voltage divider.

Potentiometer as Voltage Divider

To make a voltage divider, the Input source voltage is connected to resistance through a circuit. The two ends of resistance are fixed and one slider is present in-between the resistance which divides the given resistance into two parts R₁ and R₂. Required voltage V_out is achieved across R₂.

The below diagram will more clearly explain the situation. 

Let V_R1 and V_R2 be voltage drop across R1 and R2 respectively. By applying Kirchhoff’s voltage law, 

V_in = V_R1 + V_R2

Since V_R1 = I x R1 and V_R2 = I x R2

So, V_in = I x R1 + I x R2

V_in = I x ( R1 + R2)

I = V_in  / (R1 + R2)

Therefore, 

V_R1 = (V_in /  (R1 + R2) ) x R1

V_R1 = V_in x ( R1 / (R1 + R2))

and 

V_R2 = (V_in /  (R1 + R2) ) x R2

V_R2 = V_in x ( R2 / (R1 + R2))

The given input voltage(V_in) can be altered to the required output voltage (V_out) by sliding the variable terminal across the resistance. By moving the wiper we can change the value of R1 & R2 and hence voltage drop across R2 is also changed.

Since the value of voltage and resistance depends on each other. We know that ratio of voltages is equal to the ratio of resistances.

V_in / V_out = (R₁ + R₂ / R₂ )

So, V_out = V_in  × (R₂ / R₁ + R₂)

Generalizations: 

Case 1: If R₁ = R₂ = R,

V_out = V_in  × (R / 2R)

V_out = V_in / 2

Case 2: If R₁ << R₂

V_out ≈ V_in (R₂/R₂)

V_out = V_in

Case 3: If R₂ << R₁

V_out ≈ V_in (0/R₁)

V_out = 0

Sample Problems

Problem 1: If the current of 5A is flown through a wire in 2 sec. Find charge in wire in coulomb.

Solution:

q =  I x t

q = 5 x 2

    = 10 Coulomb

So, Charge in wire is 10 Coulomb.

Problem 2: Find the Resistance across the conductor if the current flowing is 2A and a potential difference is 5V.

Solution:

R=V/I

  = 5/2

  = 2.5 ohm

So, resistance across conductor is 2.5 ohm.

Problem 3: Find the voltage across the resistance of 3 Kilo ohm. If the current flowing through resistance is 2 A. 

Solution:

V = IR

    = 2 x 3 x 10³

= 6 K V

So, voltage across resistance is 6 KV.

Problem 4: Find the divided voltage of the potentiometer if R1 is 2 ohm, R2 is 4 ohm and the input voltage is 3V. 

Solution:

Output voltage = 3 x ( 4/ 2+4)

                        = 3 x ( 2 / 3 )

                        = 2 V

So, Output voltage of potentiometer is 2V.

Problem 5: Find the input voltage of the potentiometer if resistance is divided equally and the output voltage is 6 V

Solution:

Input voltage = 2 x 6 

                      = 12 V

So, Input voltage of potentiometer is 12V.

Problem 6: Find the voltage across R1. If source voltage is 5V , R1 = 3 ohm and R2 = 6 ohm.

Solution:

Voltage across R1 = 5 x (3 / 3 + 6)

                            = 5 x ( 3 / 9)

                            = 5 x (1 / 3)

                            = 1.67 V

So, voltage drop across R1 is 1.67V.

Problem 7: Find the output voltage if R1 = 30μ ohm and R2 = 2K ohm. The source voltage is 50V.

Solution:

Since R2>>R1, 

So, V_out is almost equal to V_in.

V_out = V_in

V_out = 50 V

So, output voltage is 50 V.