A combined circuit has two resistors in parallel (6.0 ohms and 8.0 ohms) and another in series (13.0 ohms). If the power source is 9.0 volts, what will the current be?

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A combined circuit has two resistors in parallel (6.0 ohms and 8.0 ohms) and another in series (13.0 ohms). If the power source is 9.0 volts, what will the current be?

Mathematics
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Ok the trick to this problem is first sketch the diagram....
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Pardon the terrible drawing

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Now your goal is exploit the rules to your advantage to reduce it to a simple circuit
To find the resistance of the parallel bit: \[\frac{ 1 }{ r } = \frac{ 1 }{ r_{1} } + \frac{ 1 }{ r_{2} } = \frac{ 1 }{ 6\Omega } + \frac{ 1 }{ 9\Omega } = \frac{ 5 }{ 18\Omega } \ \ \ \ \ \ \rightarrow \ \ \ \ \ \ r=\frac{ 18 }{ 5 } \Omega\] A rule of thumb is the total resistance of a parallel set of resistors will always be less than the total resistance.
Now your circuit looks like:
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To find the resistance of these simply add: Total Resistance is 13+18/5 = 16 and 3/5 ohms
By Ohms law V=IR: \[I = \frac{ V }{ R_{total} } = \frac{ 9V }{ \frac{ 83 }{ 5 } \Omega } = \frac{ 45 }{ 83 }A \approx 0.54 A\]
By the way, I need to clarify my rule of thumb. I stated it very sloppily and I hope you didn't get confused by it. What I meant to say is that the rule is a set of resisters in parallel will ALWAYS have total resistance < any of the individual resistors in the arrangement. This means if I have a bunch of resistors, but I need to build a circuit that has a lower total resistance than any of their individual resistances; than I arrange them in parallel. For example, this problem.... I needed a circuit with a resistance of lets say between 16 and 17 Ohms, but I only had some 13, 6, and 9 Ohm resistors. Clearly this is a situation perfectly suited to contain an arrangement of resistors in parallel.

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