Answer to Question #170465 in Electric Circuits for Innocent

Question

Prove that equivalent resistance R of other resistor R1 R2 R3...RN in series is given as

R= R1+R2+R3+...RN

Answer 1

By the law of the conservation of charge/the continuity of current we find that the current in each element is the same $I_1=I_2=...=I_N=I$.

As the tension $U$ is equal to the potential difference between the ends of the series of elements, we can decompose it as $U = \phi_{after}-\phi_{before}=(\phi_{after N}-\phi_{after (N-1)})+(\phi_{after (N-1)}-\phi_{after (N_2)})+...+(\phi_{after 1}-\phi_{before 1})$ where $\phi_{before/after \text{ }i}$ means the potential right before/after the i-th element. Thus, we deduce that

$U= \sum U_i$, i.e. the total tension is distributed between the elements.

Now by applying the Ohm's law to each element we have $U_i = I_i R_i$, $U=IR_{equivalent}$. Applying the two previous results we find $R_{equivalent} = \sum U_i/I = \sum R_i$