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7+ Year Member
Jun 7, 2009
  1. Medical Student
I have a 10V source, a 40 ohm and 20 ohm resistor in parallel and another 30 ohm resistor. How do i find the current in the 40 and 20 ohm resistor? I know that V= const in parallel so; 10V=I*40 I= .25 10=I*20 I=.5 Current is additive in parallel right? So total current leaving the branch is .75 amps? If that is the case then the voltage drop across the 30 ohm resistor will be; .75*30= 22.5 Volts???? this seems way wrong.. the other method is to first find total resistance, which in this case is 13.3+30= 43.3 ohms then total current is I= 10/43.3 -> 0.23 Amps. Now, this method tells me that the current leaving that 1st branch must be 0.23 A since current is constant in series in accordance to Kirchoff's law. So the voltage drop across the entire 1st branch is V=.23*13.3(equiv. resistance)--- 3.1 Volts. Voltage is constant in parallel so each resistor in the branch gets 3.1 volts. So the current in the 20 ohm resistor is 3.1/20=0.155 A current in 40 ohm resistor is 3.1/40=0.078 A Is my 2nd method the correct way of solving this?? It seems to add up properly... THANKS!
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