Question

Derive a formula to calculate effective resistance of three resistors R1, R2 and R3 when

connected in series combination.

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Answer 1

Answer:

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1)Series Connection

Derivation:

Let there be 3 resistance R1, R2, and R3 connected in series. A battery of V volts has been applied to the ends of this series combination. Now suppose the potential difference across the resistance R1 is V1, R2 is V2 and R3 is V3.

∴ V= V1+V2+V3 ...(1)

Now, suppose the total resistance of the combination be R, and the current flowing through the whole circuit be I.

So, applying Ohm's law to the whole circuit, we get :

V/I = R or V=IR ...(2)

∵ The same current is flowing through all three resistances, so by applying Ohm's law to three of them  we will get  

V1 = IR1                          

V2= IR2 and  

V3= IR3                    (3)

Now, putting (3) and (2) in equation (1), we get

IR = IR1 + IR2 + IR3

i.e IR = I ( R1 + R2 + R3 )

⇒ R = R1 + R2 + R3  

Hence its derived

2) Parallel Connection

we show 3 resistors connected 'in parallel' with one another. In this  case, the current flowing into P is divided among the 3 resistors

i = i1 + i2 + i3

However, the potential difference across any resistors  is the same, namely

i1 R1 = i2 R2 = 13 R3

These equations can be thought of as determining the currents i1, i2, i3.

Substituting, We have

i = ( V/R1 + V/R2 + V/R3 ) = V / R  

or

1/R1 + 1/R2 + 1/R3 = 1 / R.  

Similarly, For n number of resistors connected in parallel,

   The Total Equivalent resistance = 1/R1 + 1/ R2 +...+ 1/Rn = 1 / R.  

Hope this helps U ✌✌✌✌✌✌✌✌

Answer 2

Answer:

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

Similarly, for n number. The total equivalent resistance = 1/R1 + 1/R2 + 1/R3 + 1/Rn = 1/R