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Derive a formula to calculate effective resistance of three resistors R1, R2 and R3 whenconnected in series combination. |
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Answer» Answer: Hey Mate ♣♣♣ Here's Ur Answer ♣♣♣♣ 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 ✌✌✌✌✌✌✌✌ |
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