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Electrostatic physics?

  • #1
[SOLVED] electrostatic physics?

Homework Statement



a cabin contains only two small electrical appliances: a radio that requires 10 milliamperes of current at 9 volts and a clock that requires 20 milliamperes at 15 volts, a 15 volt battery with negligible internal resistance supples the electrical energy to operate the radio and the clock


Calculate the resistance and calculate the electrical energy that must be supplied by the battery to operate the circuits for 1 minute

Homework Equations



R= I/V and energy= (power)(t) and power = V x I

The Attempt at a Solution



I know that the second appliance needs 9V, so you need to drop 6V across the resistor (15-9).

does that mean i plug in 9 for V to find R? then I equals 10 milliamperes which i multiply by 10^-3 to convert to amperes

and i totally don't get the second part
 

Answers and Replies

  • #2
msg me for a picture if you think it will help you
thank you!!
 
  • #3
rock.freak667
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The resistance of what exactly are you to calculate?
 
  • #4
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A picture would help ;).

I cannot visualize circuit diagram from the question.
 
  • #5
rock freak: the directions only say calculate resistance, no additional details
 
  • #6
rock.freak667
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Post the picture
 
  • #7
Redbelly98
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R= I/V
You'd better double-check that equation.

The Attempt at a Solution



I know that the second appliance needs 9V, so you need to drop 6V across the resistor (15-9).

does that mean i plug in 9 for V to find R?
No, plug in the voltage across the resistor R.

then I equals 10 milliamperes which i multiply by 10^-3 to convert to amperes
Yes.

and i totally don't get the second part
As a start, can you calculate the power supplied by the battery?
 
  • #8
Redbelly98
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A picture would help ;).

I cannot visualize circuit diagram from the question.
A resistor and a radio are connected in series.

The resistor+radio is connected in parallel with a clock.

Then, all of that is connected to the battery.
 
  • #9
rock.freak667
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A resistor and a radio are connected in series.

The resistor+radio is connected in parallel with a clock.

Then, all of that is connected to the battery.
ah that makes sense now....
 
  • #10
my bad, it's V/I

so 15 instead of 9?


uh i really don't get how they're connected

rock freak i can't post the pic, it says im not allowed
 
  • #11
rock.freak667
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my bad, it's V/I

so 15 instead of 9?


uh i really don't get how they're connected

rock freak i can't post the pic, it says im not allowed
The radio and resistor are in parallel with the clock. since they are in parallel they have the same pd.

so the pd across the resistor+pd across radio=15
radio uses 9 => resistor has a pd of=15-9=6V
and you know that R=V/I, can you find R now?
 
  • #12
what is pd?
is I 20 milliamperes?
 
  • #13
so i multiply 20 by 10^-3
i get .02

so then i did 6/.02 and got 300 ohms
 
  • #14
Redbelly98
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my bad, it's V/I

so 15 instead of 9?
No, the number you said before:
NavitaLaVida said:
... you need to drop 6V across the resistor (15-9)
NavitaLaVida said:
what is pd?
Potential difference. It means the same thing as voltage.
 
  • #15
thanks redbelly
 
  • #16
For the resistor that is in series with a 10mA radio, to find ohms, should we use 6v/.01 or 6v/.02? My teacher gave me a whole worksheet of similar problems where you need to add a missing resistor. I don't understand whether or not we should have 20mA on the bottom. Would you use 10mA on the bottom because it needs to add to the 10mA radio?
 
Last edited:

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