Simple Elec Circuit

  • Thread starter JamesL
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  • #1
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trying to find the power delivered by a battery in a simple rc circuit.

what is the proper way to go about doing this?

i have caculated the current at the given time, and the charge on the capacitor at the given time... how do these tie into the power delivered?
 
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  • #2
chroot
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In a DC circuit, power delivered to a load at any given time is just the voltage across that load times the current through that load.

- Warren
 
  • #3
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wouldnt power delivered be different in an RC circuit?

this is a 3 part problem. in the first two parts i had to find the charge on the capacitor and the current through the resistor of this simple RC circuit (simple as in, 1 loop, 1 resistor, 1 power source, 1 cap). i asked about the current and charge on the capacitor bc in the 3rd part (which is asking what the power delivered by the battery is) it makes a point of saying:

"At 1.76 s the current in the resistor is I (from Part 2) and the charge on the capacitor is q (from Part 1)."

so that confused me.
 
  • #4
chroot
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Power delivered is as I said in my previous post; the "load" is the combination of the R and C together. The ideal cap can be assumed to not dissipate any power (its impedance is purely reactive), so all the power is dissipated by the resistor. All you need is the current through the resistor, and the voltage across the resistor. (The voltage at one end is the battery voltage. The voltage at the other is V = q/C.)

- Warren
 
  • #5
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ok. i understand that. perhaps you can look at my work and tell me if im doing something wrong?

The capactitor is 1.5 x 10^-6 Fs.

I(t) = 2.23142119 x 10^-6 A
q(t) = 4.77389 x 10^-6 C

these quantities are correct.

The voltage from the battery is 10.1 Vs... and at the other end of the resistor is V = q/C = (4.77389 x 10^-6)/(1.5 x 10^-6) = 3.1825933 V

so the voltage drop across the resistory is 6.9174066 Vs.

So power = VI = (6.9174066)(2.23142119 x 10^-6) = 1.54356 x 10^-5 Ws. I am supposed to answer in units of micro-watts, so power = 15.4356 microWs...

am i doing something wrong? bc this answer is not being counted correctly by my homework service.
 
  • #6
chroot
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I suspect you are simply using too many significant figures. Your methodology looks correct. Try submitting 15 uW.

- Warren
 
  • #7
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the online homework service requires a number of decimal places so i dont think thats it... that looks right to me too... i wonder what im doing wrong.
 
  • #8
ehild
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chroot said:
The ideal cap can be assumed to not dissipate any power (its impedance is purely reactive), so all the power is dissipated by the resistor.

- Warren
Warren, James,

This is true for alternating currents, and for the time average of the power. Here we have a transient phenomenon and instantaneous power. The energy of the source is used not only for dissipation in the resistor, but also to built up the electric field in the capacitor. The energy of the capacitor is

[tex] E_C=\frac{q^2}{2C}[/tex].

The "charging power" is

[tex] dW/dt = dE/dt = \frac{qI}{C}[/tex],

as I=dq/dt.

That has to be added to the power obtained for the resistor. But q/C=Uc, so the power of the source is Uc* I + Ur *I = Us * I.

In general, the power at time t delivered by a source is [tex]P(t)= U_s(t)*I_s(t)[/tex].

James,
The unit of power is W not Ws as you wrote. Ws is unit of energy or work, it is equal to Joule.

ehild
 

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