Question

Derive the relation between the resistors in series and parallel connection​

Answer 1

Let the current through resistor R1 be I1 and the current through R2 be I2

The voltage drop across both R1 and R2 is equal to the supply voltage V

Therefore from Ohm's Law

I1 = V / R1

and

I2 = V / R2

But from Kirchoff's Current Law, we know the current entering a node (connection point) is equal to the current leaving the node

Therefore

I = I1 + I2

Substituting the values derived for I1 and I2 gives us

I = V / R1 + V / R2

= V(1 / R1 + 1 / R2)

The lowest common denominator (LCD) of 1 / R1 and 1 / R2 is R1R2 so we can replace the expression (1 / R1 + 1 / R2) by

R2 / R1R2 + R1 / R1R2

Switching around the two fractions

= R1 / R1R2 + R2 / R1R2

and since the denominator of both fractions is the same

= (R1 + R2) / R1R2

Therefore

I = V(1 / R1 + 1 / R2) = V(R1 + R2) / R1R2

Rearranging gives us

V / I = R1R2 / (R1 + R2)

But from Ohm's Law, we know V / I = total resistance of the circuit. Let's call it Rtotal

Therefore

Rtotal = R1R2 / (R1 + R2)

So for two resistors in parallel, the combined resistance is the product of the individual resistances divided by the sum of the resistances.

Explanation:

❤️ Hope this helps you ❤️

Answer 2

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