Activity 12.6 Class 10 Science Electricity

Activity 12.6 Class 10 Science Electricity

Activity 12.6 electricityBrief procedure:

Activity 12.6 asks us to put resistors in parallel and measure the current in each of them.


Current through each resistor is different from each other.


Each of the resistors in parallel combination has the same potential difference across them; current flows as per the value of resistance in them. This happening can be explained with the example of the flow of water through different pipes. If multiple pipes are connected to the tank, water flows as per the thickness of the pipe. If the thickness of the pipe is more, more water can flow through them. Similar things happen with resistors; if the resistor is of lower resistance more current flow through them and if the resistance is high lesser current flows through them.

Here total current flow I= I1 + I2 + I3

I1:I2:I3 = 1/R1:1/R2:1/R3

This result can also be achieved through Ohm’s law, v=IR

Here potential difference across the all the resistors  are same,


V= I1*R1 = I2*R2 = I3*R3


This experiment demonstrates current in different resistors in parallel have different values.


Resistors in parallel can be assumed as a single resistor R, where R equals to

1/R = 1/R1 + 1/R2 + 1/R3


In our home electricity comes as 240v AC. Here, all the appliances act as resistors of different value and consume energy as per their requirement.


See also: Potential difference across resistors in series have are different, activity 12.5.

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