GREPhysics.NET: GR8677 #24

Jdogzz
2015-10-22 17:57:42

The problem there is that you assumed the direction of the current can be taken to flow \\\\\\\"positively\\\\\\\" no matter which way you apply the loop rule. Since the generator has the larger potential it controls the direction of the current, it must flow clockwise through the loop. So we can apply the loop rule two ways, traversing the loop clockwise or counterclockwise, and we can still subtract, as usual the potentials from components of resistance, but we will take the direction of the flow of current into account through the value of the current:\\\\r\\\\n\\\\r\\\\nMethod 1 (clockwise, so $I=10A$ ):\\\\r\\\\n $V_{gen}-IR-IR_{bat}-V_{bat}=0$ \\\\r\\\\n $120-10R-10-100=0$ \\\\r\\\\n $10-10R=0$ \\\\r\\\\n $R=1\\\\\\\\Omega$ \\\\r\\\\n\\\\r\\\\nMethod 2 (counterclockwise, so $I=-10A$ ) (this is the direction you initially used in your loop rule)\\\\r\\\\n $-V_{gen}-IR-IR_{bat}+V_{bat}=0$ \\\\r\\\\n $-120+10R+10+100=0$ \\\\r\\\\n $-10+10R=0$ \\\\r\\\\n $R=1\\\\\\\\Omega$ (the same as in method 1)\\\\r\\\\n\\\\r\\\\nIt is helpful to remember that when you do this the other way around, finding the current rather than having it provided and looking for something else), you can arbitrarily choose to do the loop rule clockwise or counterclockwise, you just have to check the resulting sign of the current at the end which tells you that you went the right way around if your answer was positive, and the opposite way if your answer was negative (though analysis ahead of time may be possible if you have a nice enough scenario where you can see which potential wins out).\\\\r\\\\n\\\\r\\\\nAlternatively, for those who would prefer not to give current a negative value, you may take the direction of the current into account by adding the potential contributions from resistances, instead of the usual subtraction, when you are crossing it opposing the flow of current as you are now acknowledging that you are going across a potential increase (so adding the potential contribution) rather than taking it at face value to be a potential decrease (subtracting the potential contribution) which you then correct later with the sign of the current:\\\\r\\\\n\\\\r\\\\nMethod 2 Alternative (counterclockwise, $I=10A$ , but resistances now being added)\\\\r\\\\n $-V_{gen}+IR+IR_{bat}+V_{bat}=0$ \\\\r\\\\n $-120+10R+10+100=0$ \\\\r\\\\n $-10+10R=0$ \\\\r\\\\n $R=1\\\\\\\\Omega$

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