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anonymous
 one year ago
A 12.0 V battery is hooked up with three resistors (R1, R2, and R3) in parallel with resistances of 2.5 Ω, 7.0 Ω, and 15.0 Ω, respectively.
Part 1: Draw a labeled circuit diagram for the circuit described using correct symbols.
Part 2: Calculate the equivalent resistance.
Part 3: Calculate the current passing through each resistor in the circuit.
***How do I do this?
anonymous
 one year ago
A 12.0 V battery is hooked up with three resistors (R1, R2, and R3) in parallel with resistances of 2.5 Ω, 7.0 Ω, and 15.0 Ω, respectively. Part 1: Draw a labeled circuit diagram for the circuit described using correct symbols. Part 2: Calculate the equivalent resistance. Part 3: Calculate the current passing through each resistor in the circuit. ***How do I do this?

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the requested circuit is: dw:1434903805092:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that is the drawing?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434903890518:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! I think so!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok yay!! so that is it for part 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ookie yay!! and so now we go to part 2? how do we do that? :/

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1part #2 the equivalent resistance is: \[\Large \begin{gathered} {R_{EQUIVALENT}} = \frac{1}{{\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}}} = \hfill \\ \hfill \\ = \frac{1}{{\frac{1}{{2.5}} + \frac{1}{7} + \frac{1}{{15}}}} = ...Ohm \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! so we get 1.640625?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay so that is our equivalent resistance? :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh okie yay!! so onto part 3!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the requested currents are: \[\Large \begin{gathered} {I_1} = \frac{{12}}{{2.5}} = ...Amps \hfill \\ \hfill \\ {I_2} = \frac{{12}}{7} = ...Amps \hfill \\ \hfill \\ {I_3} = \frac{{12}}{{15}} = ...Amps \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434904298434:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! we get these values? 4.8 1.71 0.8 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay!! and so we are finished with this problem? :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh yay!! thank you!!:) onto the next!!
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