Question

Derive the relation for effective equivalent resistance, when resistance s are connected in series.

Answer 1

Consider two resistors connected in series, as shown in Figure.It is clear that the same current $I$ flows through both resistors. For, if this were not the case, charge would build up in one or other of the resistors, which would not correspond to a steady-state situation (thus violating the fundamental assumption of this section). Suppose that the potential drop from point $B$ to point $A$ is $V$. This drop is the sum of the potential drops $V_1$ and $V_2$ across the two resistors $R_1$ and $R_2$, respectively. Thus, \begin{displaymath} V = V_1 + V_2. \end{displaymath} According to Ohm's law, the equivalent resistance $R_{\rm eq}$ between $B$ and $A$ is the ratio of the potential drop $V$ across these points and the current $I$ which flows between them. Thus, \begin{displaymath} R_{\rm eq} = \frac{V}{I} = \frac{V_1+V_2}{I} = \frac{V_1}{I} + \frac{V_2}{I}, \end{displaymath} giving \begin{displaymath} R_{\rm eq} = R_1 + R_2. \end{displaymath}