Example from Physics, 19.3 Parallel Circuits

Find the Current through a Complex Resistor Circuit

The battery in the circuit below has a voltage rating of 10 V. What current flows through the circuit and in what direction?

A circuit diagram showing six resistors, some in series and others parallel.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Apply the Strategy for finding equivalent resistance to replace all the resistors with a single equivalent resistance, then use Ohm’s law to find the current through the equivalent resistor.
Solution
The resistor combination $R_{4}$ and $R_{5}$ can be reduced to an equivalent resistance of

$R_{P1} = \frac{1}{1 / R_{4} + 1 / R_{5}} = \frac{1}{1 / 45 Ω + 1 / 60 Ω} = 25.71 Ω R.$
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Replacing $R_{4}$ and $R_{5}$ with this equivalent resistance gives the circuit below.

We now replace the two upper resistors $R_{2}$ and $R_{3}$ by the equivalent resistor $R_{S 1}$ and the two lower resistors $R_{P 1}$ and $R_{6}$ by their equivalent resistor $R_{S2}$ . These resistors are in series, so we add them together to find the equivalent resistance.

$\begin{matrix} R_{S1} & = & R_{2} + R_{3} & = & 50 Ω + 30 Ω & = & 80 Ω \\ R_{S2} & = & R_{P1} + R_{6} & = & 25.71 Ω + 20 Ω & = & 45.71 Ω \end{matrix}$
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Replacing the relevant resistors with their equivalent resistor gives the circuit below.

Now replace the two resistors $R_{S1} and R_{S2}$ , which are in parallel, with their equivalent resistor $R_{P2}$ . The resistance of $R_{P2}$ is

$R_{P2} = \frac{1}{1 / R_{S1} + 1 / R_{S2}} = \frac{1}{1 / 80 Ω + 1 / 45.71 Ω} = 29.09 Ω.$
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Updating the circuit diagram by replacing $R_{S1} and R_{S2}$ with this equivalent resistance gives the circuit below.

Finally, we combine resistors $R_{1} and R_{P2}$ , which are in series. The equivalent resistance is $R_{S3} = R_{1} + R_{P2} = 75 Ω + 29.09 Ω = 104.09 Ω .$ The final circuit is shown below.

We now use Ohm’s law to find the current through the circuit.

$\begin{matrix} V & = & I R_{S3} \\ I & = & \frac{V}{R_{S3}} & = & \frac{10 V}{104.09 Ω} & = & 0.096 A \end{matrix}$
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The current goes from the positive terminal of the battery to the negative terminal of the battery, so it flows clockwise in this circuit.

0 of 2 steps open

How did it go?

Strategy

Solution

Strategy

Solution