Voltage and resistance in series connection

Question

In a series connection with n elements it is true that (voltage):

$$V = V_1 + V_2 + ... +V_n$$

and (resistance):

$$R = R_1 + R_2 + ... +R_n$$

If I know one of these I can infer the other. But is it possible to prove any of them without the other?

Do you know the definition of $U$ in terms of the more elementary electric field?
If you now imagine having two islands in series. You need the energy (per charge) $U_1$ to bring a unit charge in the first island and the energy $U_2$ to bring it from the first to the second. Then because energy itself is additive you need the energy $U_1+U_2$ to bring it directly to the second island (so the first equation is more elementary and derives from the additivity of energy).

Alfred Centauri · Answer 1 · 2012-08-07 19:22:40Z

But is it possible to prove any of them without the other?

(1) By KVL, the voltage across the N resistors is:

$V = V_{R_1} + V_{R_2} + ... + V_{R_N}$

(2) For a series connection, by definition, there is only one current, $I$.

By Ohm's Law, the voltage across any of the series resistors is:

$V_{R_n} = I \cdot R_n$

By KVL:

$V = I \cdot R_1 + I \cdot R_2 + ... + I \cdot R_N = I \cdot (R_1 + R_2 + ... R_N) = I \cdot R$

$R = R_1 + R_2 + ... + R_N$

Lucy Brennan · Answer 2 · 2012-08-08 05:10:35Z

Here is a very slow derivation of $U = U_1 + ... + U_n$

The energy transformed in both of them must equal the sum of the energy transformed in each of them:

$P = P_1 + P_2 + ... + P_n$

According to the definition of electric power:

$P = IV$

By combining the two:

$I*V = I_1*V_1 + I_2*V_2 + ... + I_n*V_n$

But since the current is the same at any point in a series connection, $I$ can be crossed out on both sides.

$V = V_1 + V_2 + ... + V_n$

2 Answers