A 15.0 Ω resistor is hooked up in series with a 12.0 Ω resistor followed by a 9.0 Ω resistor. The circuit is powered by a 12.0 V battery. a.) Draw a labeled circuit diagram for the circuit described. b.) Calculate the equivalent resistance. c.) Calculate the voltage drop across each resistor in the circuit. **How would this work? Thank you!! :)
please wait a moment
okie!:)
here is your circuit: |dw:1432233280169:dw|
ohh okie :) whoahh so is that all that is needed for part a? :O
yes!
yay!! :) how do we do part b? :/
It is simple the equivalent resistance , is given by the sum of your resiatances, namely: \[{R_{{\text{eq}}}} = {R_1} + {R_2} + {R_3} = ...?\]
ohh so 9+12+9=30? so part b equivalent resistance is 30 ohm?
I got 15+9+12=36
oh oops!! yes sorry, i do not know why i put 9 twice!! so yes, it would be 36 ohm is the eq. resistance?
yes!
yay!! okay, so last part? how do we do part c? :O
we have to compute the current of our circuit. that current intensity, is given by the subsequent formula: \[I = \frac{V}{{{R_{{\text{eq}}}}}} = ...?\]
so 12/36? and we get 0.3333.... ? that is the voltage drop? :/
that's right! It is the current, no the voltage drop
ohh okay!! but part c asks for calculation of voltage drop across each resistor... how can we find that?
the requested voltage drops, can be computed using the Ohm's law, as below: \[\begin{gathered} {V_1} = {R_1}I = ...? \hfill \\ {V_2} = {R_2}I = ...? \hfill \\ {V_3} = {R_3}I = ...? \hfill \\ \end{gathered} \]
ohh so 15*0.33 = 4.95; 9*0.33=2.97 and 12 * 0.33=3.96 those are at the voltage drops?
that's right! Since I = 1/3, w better is: \[\begin{gathered} {V_1} = {R_1}I = \frac{{15}}{3} = 5volts \hfill \\ {V_2} = {R_2}I = \frac{{12}}{3} = 4volts \hfill \\ {V_3} = {R_3}I = \frac{9}{3} = 3volts \hfill \\ \end{gathered} \]
ohh okay!!! yay!! thank you!! is this problem finished? :O
oops.. Since I = 1/3, better* is:
yes! we have finished!
yay!! thank you so much!! :D
thank you!! :)
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