I have a circuit with a constanst voltage source a resistor in parallel to the voltage source and a capacitor in parallel with the resistor and a resister perpendicular to the voltage source and the other resistor. I've calculated the resistance but it differs from those that I've calculated through Ohms laws calculates I've found on the web. Is the variation because Ohms law wont work when there is a capacitor in the circuit?
For example I used this calculator: http://www.anderson-bolds.com/calculator.htm
2007-07-17 06:54:23 · 5 answers · asked by kells_2010 1 in Science & Mathematics ➔ Engineering
I'm using PSPICE so any suggestions on how I can find the resistance of a circuit that contains one capacitor and two resisters.
2007-07-19 07:46:49 · update #1
The capacitor should show low resistance until it's charged by the voltmeter, battery. Then all the current will go through the resistor. It should be pretty close to the right resistance but there may be other things in the circuit (Including the power supply). Try disconnecting the P.S.and re-take your reading. If it's still off, remove the capacitor.
Taking readings in a circuit is difficult because of all the surrounding circuitry. Are you taking voltage readings and calculating the resistance or directly measuring ohms?
BTW, thanks for the website with the calculators !!
2007-07-17 07:08:46 · answer #1 · answered by Dan Bueno 4 · 0⤊ 0⤋
correct. the capacitor is a frequency dependant component. it's 'resistance' (actually impedance) changes if the voltage supplied changes.
for a constant voltage situation there is a charge time during which there is some current through the capaitor, but the amount of current decreases as time passes. after enough time has passed the current in that leg stops and the only resistance is the one r in paralle to the voltage source.
the time required depends on the values of the resistor and capacitor in series.
2007-07-17 07:08:51 · answer #2 · answered by Piglet O 6 · 1⤊ 0⤋
A multimeter set to amps is an ammeter. A multimeter set to volts is a voltmeter. A multimeter set to ohms reasures resistance by making use of making use of a widely used voltage to the burden and measuring the present by it. precisely what voltage is used relies upon on the multimeter, and what variety it it set to. A digital multimeter does the arithhmetic for you and reflects the respond. you does not connect 2 ammeters in sequence - they might desire to easily reveal the comparable fee. you does not connect a voltmeter and ammeter in sequence - an acceptable voltmeter has countless impedance, so might study the completed volts on a similar time as the ammeter study 0. For a genuine voltmeter, the ammeter shows the small modern drawn by making use of the voltmeter, which isn't functional. What you may routinely do, is connect a power furnish and ammeter in sequence for the duration of a load, then connect a voltmeter in simple terms around the burden. then you definitely've the present during the burden, and the voltage for the duration of it, and can use ohm's regulation to discover the resistance. this gadget is honestly greater precise than making use of a multimeter for small resistances, because of the fact it avoids blunders because of the touch resistance. in case you have 2 multimeters, i assume you will desire to set one to resistance and one to voltage and connect the two around the burden. then you definitely might know the resistance, and the voltage utilized by making use of the meter in resistance mode, so would desire to artwork out the present.
2016-12-14 11:33:00 · answer #3 · answered by lunger 4 · 0⤊ 0⤋
This is an AC circuit, you would be calculating the Impedance (Z), not the resistance. Where Z = V/I. You'll need to calculate the current drawn by the capacitor (Xc component).
2007-07-24 22:50:26 · answer #4 · answered by Kitty L 1 · 0⤊ 0⤋
ohms law will consider the capacitor to be an open circuit. In that sense you can still calculate the total resistanc.e
2007-07-17 07:05:44 · answer #5 · answered by Anonymous · 0⤊ 0⤋