Derive the relation `R = R_1 + R_2 + R_3` when three resistors `R_1, R_2 and R_3` are connected in series in an electrical circuit.

Explanation

Let the potential difference across the resistors `R_1, R_2 and R_3` are, `V_1, V_2 and V_3` respectively. In the electric circuit shown above, let I be the current through the circuit. In this case, the current through each resistor is also I and the potential difference across a combination of resistors is equal to the sum of potential differences across this individual resistors. That is
`V = V_1 + V_2 + V_3 " "....(1)`
If the resistors `R_1, R_2` and `R_3` are replaced by an equivalent single resistor R, such that the potential difference V across it, and current I through the circuit remain the same, then, applying the Ohm.s law to the entire circuit and three. resistors separately, we have
`V = IR " " ...(2)`
`V_1 = IR_1 " " ....(3)`
`V_2 =IR_2`
`V_3 = IR_3 " " .....(4)`
By adding (3), (4) and (5), we get
`V_1 + V_2 + V_3 = IR_1 + IR_2 + IR_3`
or `V_1 + V_2 + V_3(R_1 + R_2 + R_3)` .
By using equation (1)
`V = I(R_1 + R_2 + R_3)`
By using equation (2)
`IR = I(R_1 + R_2 + R_3)`.
or `R = R_1 + R_1 + R_3`