AIM

To determine the equivalent resistance of two resistors when connected in parallel.

MATERIALS REQUIRED

Two standard resistance coils, ammeter, voltmeter, one-way plug key, a low resistance rheostat, connecting wires, battery or battery eliminator.

THEORY

An arrangement in which resistors are connected between two common ends of the circuit in such a way that the potential difference across each resistance is equal to the applied voltage, then such an arrangement is called the parallel combination

As shown in the figure, two resistances R₁ and R₂ are connected between two points A and B  in the parallel combination.

Let the potential applied by the dc source to this combination be V. let I₁ and I₂ be the current measured by ammeters, connected in series with the resistor R₁ and R₂ respectively, then

I = I₁+I₂

According to ohms law, I₁ = V/R₁ and I₂ = V/R₂

If R_p, is the equivalent resistance of the given parallel combination having the same potential difference a the applied potential, then

or

Circuit Diagram

Apparatus Arrangement Diagram:

Connect all the resistors in parallel combination between the two terminals of the voltmeter as shown in the figure given below.

PROCEDURE

1. Connect the circuit as shown in the circuit diagram with one of the unknown resistors.

2. By using Ohm’s law, find the value of each resistance, Let it be R₁ or R₂.

3. Connect the given resistors in parallel combination between the two terminals of the voltmeter as shown in the circuit diagram above.

4. Plug in the key and take the readings of ammeter and voltmeter.

5. Repeat the step 5 for three times by changing the position of the sliding contact of the rheostat.

OBSERVATION TABLE

Resistor used	No. of observations	Voltmeter reading ‘V’ (in volt)	Ammeter reading ‘I’ (in ampere)	(in ohm)	Mean value of resistance (in ohm)
R1	(a) (b) (c)	0.20 0.45 0.60	0.1 0.2 0.3	2.00 2.25 2.00	= 6.25 Ω
R2	(a) (b) (c)	0.15 0.45 0.65	0.1 0.2 0.3	1.50 2.25 2.17	= 5.92 Ω
	(a) (b) (c)	0.10 0.25 0.30	0.1 0.2 0.3	1.00 1.25 1.00	= 1.083 Ω

Observation:

1. Least count of ammeter:

The image of the ammeter is attached here:

The range of the ammeter = 500 - 0 mA = 500 mA = 0.5 A

The number of divisions in between two consecutive values = 10

Therefore, the least count =

= 0.01 A

2. Zero error of ammeter:

The needle of the ammeter points towards zero of the main scale of the ammeter.

3. Least count of voltmeter:

The range of voltmeter = 2 - 0 V = 2 V.

The number of division in small scale between two division on the main scale

= 10

Therefore, the least count

4. Zero error of voltmeter = 0 V.

CALCULATIONS

1. Mean value of R₁ = 2.083 Ω

2. Mean value of R₂ = 1.973 Ω

Equivalent value of parallel combination:

(a) by calculation, R = 1 Ω

RESULT

1. The equivalent resistance of parallel combination = 1.08 Ω.

2. There is a direct agreement between the calculated value and the experimental value.

Hence, is verified.

PERCENTAGE ERROR

Percentage error

Percentage error = 8%

Precautions:

1. All the connection should be tight and properly done as per circuit diagram.

2. Take out the plug from the plug key in between the two observations.

3. A low resistance rheostat should be used in the circuit to obtain large variation in current.

4. A thick copper connecting wires should be used in the circuit.

5. The positive terminal of the ammeter and voltmeter must be connected with the positive terminal of the battery or battery eliminator.