Physics Bootcamp

Subsection 36.2.1 Current and Voltages in Series

In Figure 36.2.1, the resitors in series are connected to a voltage source such that we set up a potential differece across the two resistors.

But since we do not know the potential drop across either of the two resitors, we cannot apply Ohm's law to either resitor to find the current through them. If we apply Ohm's law to each of the two resitors we will get the following equations with unknown \(\phi_c\) as well as \(I\text{.}\)

We can think of finding \(I\) as an algebra problem for these two equations. You can eliminate \(\phi_c\) by adding the two equations and solve for \(I\text{.}\)

Using this in the two equations above, we can immediately find the voltage drops across each resitor.

From these you can see that total voltage drop \(V\) has been proportionaltely shared by the two resistors in series. We say that resitors in series “divide-up” the total voltage acoss them in proportion of their own resitance - the resitor with higher resitance has the higher proportion of voltage across it.

An open circuit with no connection, say between \(b\) and \(c\) in Figure 36.2.1 will have \(R_2=\infty\text{,}\) and all voltage will drop across such an open line. We call such places in a circuit to be open circuit.

Subsection 36.2.2 Equivalent Resistance of Resitors in Series

It is instructive to think of resistors in series conceptually, as well, which we do next. To find the common current \(I\) through resistors in series, we will combine them into one resistor, to which we can apply Ohm's law. For instance, if resistor \(R_1\) is a \(5\text{-m}\) copper wire and \(R_2\) is a \(3\text{-m}\) copper wire, then we can replace them with \(5+3=8\text{-m}\) wire of same thickness. That means, the net resitance of two resistors in series is

These arguments for two resitors in series can be easily extend to the case of \(N\) resistors in series, giving us

We say that between points \(a\) and \(b\text{,}\) in Figure 36.2.1 the two resistors present an equivalent resistance, \(R_\text{equiv}\text{,}\)

We may think of an equivalent circuit that has only one resitor, with resistance \(R_\text{equiv} \) rather than two resistors of the original circuit, without affecting the original circuit outside of points \(a\) and \(b\), as indicated in Figure 36.2.2.

Now, we can apply Ohm's law for this equivalent resistor to obtain current \(I\) as

where I have replaced \(\phi_a-\phi_b\) by \(V\) as we noted in Eq. (36.2.1).

Subsection 36.2.3 Potential Drops Across Each Resistor in Series

Now, we can use Ohm's law on each of the resistor, as in Eqs. (36.2.2) and (36.2.3), to find potential drops across each.

An important conceptual lesson to note here is that total potential drop across the two resistors, \(\Delta\phi = \phi_a - \phi_b\equiv V\text{,}\) is actually a sum of the potential drops \(\Delta\phi_1 = \phi_a - \phi_c\equiv V_1\) and \(\Delta\phi_2 = \phi_c - \phi_b\equiv V_2\) across the two resitors in series.

\begin{equation*} V = V_1 + V_2. \end{equation*}

We say that original voltage across the two has been divided up between the two resitors. We say that resitors in series is a volatage-divider circuit! Eqs. (36.2.7) and (36.2.8) tell us that the total voltage drop \(V\) has been proportionaltely shared by the two resistors in series. We say that resitors in series “divide-up” the total voltage acoss them in proportion of their own resitance - the resitor with higher resitance has the higher proportion of voltage across it.