The algebraic sum of the products of the resistances of the conductors and the currents in them in a closed loop is equal to the total emf available in that loop.
No, V is actually the voltage "drop" across the resistor. Eg, if 1K is tied between 5V and 3.3V, the current I = Vdrop / R = (5V - 3.3V) / 1K = 1.7 mA, and not 5 mA or 3.3 mA.In your example, you mean a 1K(ohm) resistor, right? (just making sure--I'm really new, if you haven't figured it out yet. )So, in this LED (for example), it says "forward voltage drop" is 1.8V-2.2V (DC). Let's say I have a circuit that has 5V---LED---1K(ohm)---Gnd. I want to find the current. I = Vdrop / R = 2.2V / 1K = 2.2mA.Right? Or am I still confused... :/
or 1 / .2 = a 5 ? resistor?