## anonymous one year ago 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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1. anonymous

Ok the trick to this problem is first sketch the diagram....

2. anonymous

|dw:1440736937966:dw|

3. anonymous

Pardon the terrible drawing

4. anonymous

Now your goal is exploit the rules to your advantage to reduce it to a simple circuit

5. anonymous

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.

6. anonymous

7. anonymous

|dw:1440737490303:dw|

8. anonymous

To find the resistance of these simply add: Total Resistance is 13+18/5 = 16 and 3/5 ohms

9. anonymous

By Ohms law V=IR: $I = \frac{ V }{ R_{total} } = \frac{ 9V }{ \frac{ 83 }{ 5 } \Omega } = \frac{ 45 }{ 83 }A \approx 0.54 A$

10. anonymous

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.