lab activity-describe the connection of cells in parallel and in series
Answers
Explanation:
Aim
To determine the equivalent resistance of two resistors when connected in parallel.
Theory
When the resistors are connected in parallel with a combination of cells or battery, in such case the total current I, is equal to the sum of the separate value of current through each branch of the combination.
i.e., I=I1+I2+I3+…..
In the above circuit let R be the equivalent resistance of the parallel combination of resistors.
∴ By applying Ohm’s law we have
I=V/Rp …(1)
On applying Ohm’s law to each resistor we get

When resistors are connected in parallel combination the total resistance is reciprocal sum of the individual resistances.
i.e., 1/Rp = (1/R1) + (1/R2)
Current is constant in series circuit. Hence, we cannot connect bulb and room heater in series because their current requirement is different.
Hence such devices are connected in parallel so that the current is divided through the different electrical gadgets.
The total current is always decreased when resistors are connected in parallel.
When the resistors are connected in parallel then the equivalent resistance of the parallel combination of the resistors is always low.
Materials Required
A battery, a plug key, connecting wires, an ammeter, a voltmeter, rheostat, a piece of sand paper and two resistors of different value.
Procedure
Keep the key off and make all the connections as shown in the given figure I.
When the circuit is connected appropriately insert the key.
Note three readings of ammeter and voltmeter for the resistors R1 and R2 separately.
Now connect the circuit as shown in figure II below.
The resistors are connected in parallel and voltmeter is also connected in parallel.
Use the rheostat and record three different readings of ammeter and voltmeter.
Remove the key.
Do the calculations from the observation table.
Circuit Diagrams

Observation Table For Resistance In Parallel
Resistor
Used
No. of
Observations
Voltmeter Reading in Volts (V)Ammeter Reading in Ampere (I)R=V/I
(in Ohm)
Mean Value of Resistance (Ohm)R1
(Ist Resistor)
(a)0.010.011R1 = 1 Ohm(b)0.020.021(c)0.040.041R2
(2nd Resistor)
(a)0.020.012R2 = 2 Ohm(b)0.060.032(c)0.080.0421/Rp= (1/R1)+ (1/R2)
(ParallelCombination)
(a)0.0260.040.67Rp = 0.67 Ohm
1/Rp =1.5 Ohm
Result
The calculated value of 1/Rp = (1/R1) + (1/R2) = 1.5 Ω
The experimental value of 1/Rp = 1.5 Ω
The equivalent resistance (Rp) is less than the individual resistance (R1 or R2)
Precautions
The connecting wires should be thick copper wires and the insulation of their ends should be removed using the sand paper.
Connections should be tight otherwise some external resistance may introduce in the circuit.
Connections should be made as per the circuit.
The ammeter should be connected in series with the resistor such that the current enters at the positive terminal and leaves at the negative terminal of the ammeter.
Voltmeter should always be connected in parallel to resistor.
Calculate the least count of voltmeter and ammeter correctly.
The pointers of the ammeter and voltmeter should be at zero mark when no current flows through the circuit.
Current should be passed through the circuit for a short time while taking observations; otherwise current would cause unnecessary heating in the circuit. Heating may change the resistance of resistors.
I hope it will helpful to you give thanks to my answer nd follow me plz and pliss Mark my answer brainliest