EPT![[image loading]](http://i207.photobucket.com/albums/bb245/icydude21/circuit.jpg)
so the questions are
1) Simplify circuit to find the Current (A)
2) What is the voltage across the capacitor
Thanks
Easy Physics Question
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GrayArea
United States872 Posts
so the questions are 1) Simplify circuit to find the Current (A) 2) What is the voltage across the capacitor Thanks | ||
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Mooga
United States575 Posts
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Mooga
United States575 Posts
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Mooga
United States575 Posts
The total resistance is 4 ohms, thus the amperage through the circuit is 3 amps. Which gives you 6 volts across the capacitor. | ||
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GrayArea
United States872 Posts
On May 23 2009 15:23 Mooga wrote: I'm not sure why your circuit has a capacitor in there if there is not switch in the circuit. The total resistance is 4 ohms, thus the amperage through the circuit is 3 amps. Which gives you 6 volts across the capacitor. Can you show the calculations? I think my answerbook has a typo. | ||
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Pseudo_Utopia
Canada827 Posts
But in the DC case, you'll indeed have a 4 amp resistance but only after the capacitor is fully charged. At first, the effective resistance of the parallel branches will be 0, because Q = CV implies that when there is no charge on the capacitor (Q=0), then V=0 and so the capacitor "acts like a wire". Hope this helps ^^ | ||
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Mooga
United States575 Posts
I = V/R, I = 12/4 = 3 amps V = IR = 3 * 2 = 6 volts Assuming that the battery is a DC source, then my answer should be right because the capacitor should behave as an open circuit. Edit: And I'm assuming that it is at steady-state. | ||
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GrayArea
United States872 Posts
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Mooga
United States575 Posts
On May 23 2009 15:43 GrayArea wrote: Thanks Mooga, really appreciate the help. I think they wrote the answer to a wrong circuit in the back for this problem. No problem, good luck with physics. | ||
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Saracen
United States5139 Posts
above answers are true for t=much at t=0 the capacitor has stored no charge and acts like a wire, so Rtot = 2 ohms Qcap = 0 C Vcap = 0 V I = 6 A Vr1 (top resistor) = 0 V Vr2 (bottom resistor) = 12 V | ||
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fight_or_flight
United States3988 Posts
On May 23 2009 16:02 Saracen wrote: at t=0 the capacitor has stored no charge and acts like a wire, so hm, but if you are solving for the transient then I don't see how you can assume this. You will have an equation with 2 variables: time and initial charge/voltage on the capacitor | ||
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