Question

Derive laws of resistance in series so as to explain equivalent resistance of circuit?​

Answer 1

Answer:

In a series circuit, the equivalent resistance is the algebraic sum of the resistances. The current through the circuit can be found from Ohm's law and is equal to the voltage divided by the equivalent resistance. The potential drop across each resistor can be found using Ohm's law.

Answer 2

Answer:

Resistors are the basic components of any electrical or electronic circuit. Often resistors are found in large numbers irrespective of the size of the circuit. Resistors can be connected in series or in parallel or a combination of both. In order to reduce the complications of different combinations of resistors, some rules are to be followed.

Two resistors are said to be in series when same current flows through them. Resistors in series can be replaced by a single resistor. All the resistors follow the basic laws like Ohm’s law and Kirchhoff’s current law irrespective of their combination and complexity.

A set of resistors are said to be in series when they are connected back to back in a single line. The same current will flow through all the resistors. Resistors in series are said to have common current.

Derivation of expression for equivalent resistance of three resistors connected in series.-

Total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors.  

 $\sf{{V=}{V_1}+{V_2}+{V_3}}$

Let I be the current in the circuit. The current through each resistor is also I. It is possible to replace the three resistors joined in series by an equivalent resistor of resistance R. Applying Ohm's law,  

V = IR

On applying Ohm's law to the three resistors respectively we further have  

 $\sf{{V_1}={I{R_1}}}$

 $\sf{{V_2}={I{R_2}}}$  

 $\sf{{V_3}={I{R_3}}}$

But              

$\sf{{V={V_1}+{V_2}+{V_3}}$

$\sf{{IR}={I{R_1}}+{I{R_2}}+{I{R_3}}$

$\sf{{R}={R_1}+{R_2}+{R_3}}$