1.2: Examples- Voltage and Current Dividers

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James Kirtley
Massachusetts Institute of Technology via MIT OpenCourseWare

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Figure 6 may be used as an example to show how we use all of this. See that it has one loop and three nodes. Around the loop, KVL is:

\(\ V_{s}-v_{1}-v_{2}=0\)

At the upper right- hand node, we have, by KCL:

\(\ i_{1}-i_{2}=0\)

The constitutive relations imposed by the resistances are:

\(\ \begin{array}{l}
v_{1}=R_{1} i_{1} \\
v_{2}=R_{2} i_{2}
\end{array}\)

Combining these, we find that:

\(\ V_{s}=\left(R_{1}+R_{2}\right) i_{1}\)

We may solve for the voltage across, say, R₂, to obtain the so-called voltage divider relationship:

\(\ v_{2}=V_{s} \frac{R_{2}}{R_{1}+R_{2}}\label{4}\)

Screen Shot 2021-07-18 at 8.30.03 PM.png

A second example is illustrated by Figure 7. Here, KCL at the top node yields:

\(\ I_{s}-i_{1}-i_{2}=0\)

And KVL, written around the loop that has the two resistances, is:

\(\ R_{1} i_{1}-R_{2} i_{2}=0\)

Combining these together, we have the current divider relationship:

\[\ i_{2}=I_{s} \frac{R_{1}}{R_{1}+R_{2}}\label{5} \]

Once we have derived the voltage and current divider relationships, we can use them as part of our “intellectual toolkit”, because they will always be true.

Screen Shot 2021-07-19 at 10.02.08 AM.png