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DODGEVIPER13 said:Yes this is the correct circuit.
No, that's the potential drop across the 5Ω resistor. The current is 13A and the resistance is 5Ω, so the potential drop is I*R = 13*5 = 65V. If the potential drop across the resistor is 65V, and the left end of the resistor is at 18V, what must be the potential at its other end (The Vx end)?DODGEVIPER13 said:ok well Vx is any easy one its 65 volts because V=ir so 13(5)=65.
An ideal voltage source has no resistance, and by your description you seem to be adding resistances and voltages together, which is not correct. So no, your approach is not correct.RA is a bit more difficult for me so I assume it would be this way first I find for the resistance of the voltage source 65/13=5 ohms so 0=-5(ohms from voltage source) - 5(ohms from resistor)+6(3)-8RA. solving for RA i get 1 ohm which is correct but did I do it right?
What direction is the 13A current flowing? So what is the direction of the potential drop across the 5Ω resistor?DODGEVIPER13 said:So Vx=47 volts because 65V-18V=47V right.
Sum the known currents flowing into the node. Subtract the known currents flowing out. What's left?0=iA-13+3, so the current through RA is 10 A that doesn't semm right ugh??
It's negative for the same reason that you've made ix negative: Because it is flowing out of the node, and you've assumed that currents flowing into the node are positive.DODGEVIPER13 said:0=-iA+13+8-3 so iA=18A then my potential is 18v= 18RA so RA=1 ohm. But why is iA negative??
as to the Vx question it is going to the left across the 5 ohm resistor right so would that be -65+18=-47 volts
DODGEVIPER13 said:well since I have done all else the answer must be 18+65=83 volts but I would assume the reason is the plus is on the right and minus on the left of the resistor. I guess this would be because the Vx positive side is up so therefore the resistors pos side must be on the right.
To use KCL (Kirchhoff's Current Law) to solve for resistance in a circuit, you must first draw a circuit diagram and label all the known values for current and voltage. Then, apply KCL by writing an equation that states the sum of all currents entering a junction must equal the sum of all currents leaving the junction. Use this equation to solve for the unknown resistance value.
The formula for solving for resistance using KCL is R = V/I, where R is the resistance in ohms, V is the voltage in volts, and I is the current in amperes. This formula is based on Ohm's Law, which states that resistance is equal to voltage divided by current.
Yes, KCL can be used to solve for resistance in any type of circuit, as long as the circuit is in a steady state and follows the principles of Kirchhoff's laws. This includes series and parallel circuits, as well as more complex circuits with multiple loops and branches.
To check your solution for resistance using KCL, you can use a multimeter to measure the voltage and current at different points in the circuit and compare them to the values you calculated using KCL. If they are within a reasonable margin of error, your solution is likely correct.
One limitation of using KCL to solve for resistance is that it assumes the circuit is in a steady state, meaning that the values for voltage and current are not changing over time. If the circuit is not in a steady state, KCL may not provide an accurate solution. Additionally, KCL does not take into account the internal resistance of voltage sources, which can affect the overall resistance in the circuit.