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 Mar 2nd 2011, 07:07 AM #1 Junior Member   Join Date: Sep 2010 Posts: 24 Superpostion Principle Looking for someone to look over my work as I am having a hard time with this unit. Thank you in advance! Using superposition find a) the current through the 10-olm resistor and b) its direction, up or down, and c) the power dissipated in it. a) can someone give me a hint as to what I need to do here?
Mar 2nd 2011, 07:29 AM   #2
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 Originally Posted by quantum_enhanced Looking for someone to look over my work as I am having a hard time with this unit. Thank you in advance! Using superposition find a) the current through the 10-olm resistor and b) its direction, up or down, and c) the power dissipated in it. a) can someone give me a hint as to what I need to do here?
I think I know what I need to do. I need to convert it to a simpler circuit by combining the current source with the 10-olm resistor. E_s=(9 A)(10\Omega)=90 V and now everything is in series with a 90 V power source. Therefore, the current through the 10-olm resistor is, I=(90 V + 18 V)\(18\Omega)= 6 A

b) its direction would be up.

c)P=(6 A)^2 * 10 = 360 J

Last edited by quantum_enhanced; Mar 2nd 2011 at 07:34 AM.

 Mar 2nd 2011, 08:34 AM #3 Senior Member   Join Date: Dec 2008 Location: Udupi, Karnataka, India Posts: 292 You can solve this in two steps. i) Short circuit the voltage source. Find the current through the resistances by using branch current formula i.e. I2 = I*R1/(R1+R2). Note the direction of the current. ii) Open the current source. Find the current through the resistances by ohms law. Note the direction of the current. Next find the net current through R1.
 Mar 2nd 2011, 10:16 AM #4 Junior Member   Join Date: Sep 2010 Posts: 24 I'm assuming for the second step we have to short circuit the current source, correct? For example, when I short circuit the voltage source, I have 9 A of current. Therefore, through the 10 olm resistor I have: I= (9 A)(10 olm)/(18 olm)=5 olm down Then I short circuit the current source, and I have: I=18V/18olm=1 A up, Therefore, there is a total current flowing through the 10 olm resistor of 5-1=4 olm.. Does that look right?
 Mar 2nd 2011, 06:00 PM #5 Senior Member   Join Date: Dec 2008 Location: Udupi, Karnataka, India Posts: 292 No. The ideal current source has infinite internal resistance and ideal voltage source has zero internal resistance. If you short circuit the current source, that path will have zero resistance and current will be infinite. The voltage source will be crashed.

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