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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?
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok the trick to this problem is first sketch the diagram....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440736937966:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Pardon the terrible drawing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now your goal is exploit the rules to your advantage to reduce it to a simple circuit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now your circuit looks like:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440737490303:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To find the resistance of these simply add: Total Resistance is 13+18/5 = 16 and 3/5 ohms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By Ohms law V=IR: \[I = \frac{ V }{ R_{total} } = \frac{ 9V }{ \frac{ 83 }{ 5 } \Omega } = \frac{ 45 }{ 83 }A \approx 0.54 A\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By 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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