Common problems in introductory physics classes ask for the voltage drop across resistors. The fundamental configurations examined in introductory courses are resistors in series and resistors in parallel. Lessons on inductors (coils) and the voltage drops they cause usually follow the lessons on resistors. Inductors also cause voltage drops in a circuit, just as resistors do, but the process is different. The voltage drops are not constant like in most resistors, but the result with respect to voltage drop is effectively the same.

Calculate the voltage drop across a series of resistors by summing up their resistances and multiplying by the current through them. The same current will run through each one, by the law of conservation of mass and the law of conservation of charge, so if you know the current through one, you know the current through them all. For example, if the resistances of three resistors in series is 3 ohms, 2 ohms and 4 ohms, and the current through one of them is known to be 2 amps, then the voltage drop through the length of them is (3+2+4)*2 = 18 volts, assuming resistances in the wire between the resistors is negligible or included in the three resistance measurements already.

Calculate the voltage drop across series in parallel with the formula 1/R(total) = 1/R1 + 1/R2, where R1 and R2 are the resistances of the two parallel resistors. R(total) is the equivalent resistance of the two parallel resistors. In other words, if you removed the parallel structure in the electrical circuit and replaced it with a resistor with resistance R(total), the voltage drop through that stretch would be the same. This is a result of what is called Kirchhoff's loop rule. Any closed loop around the circuit must sum to zero potential drop, if you include any voltage sources as having opposite-signed voltage drop from the resistors' voltage drop. Then you will just use the current before or after the parallel section to calculate the voltage drop by V=i*R(total). For example, if R1=1 ohm and R2=2 ohms, then 1/R(total) = 1/1 + 1/2 = 1.5, so R(total) = 2/3. If the current at any point in the same circuit is, say, 3 amps, then the voltage drop through each of the resistors must be 3 x 2/3 = 2 volts.

Calculate the voltage drop across an inductor (coil) by multiplying the inductor's inductance by the rate of change of the current through it. Inductance is measured in henries (H). So if a coil of inductance 5 Henries has a current increasing in it by 2 Coulombs per second at that instant in time, then the induced electromotive force (emf) that opposed the voltage driving the current is 10 volts at that instant in time. This induced emf is equivalent to a voltage drop in a resistor, because it reduces the voltage as the electrons traverse the electrical circuit, just as a resistor would reduce the voltage.

Calculate the voltage drop (actually a voltage gain) across the secondary coil of a transformer from the voltage drop across the transformer's primary coil by using the formula V1/n1 = V2/n2, where n1 is the number of turns the primary coil makes around the transformer and n2 is the number of turns the secondary coil makes around the transformer. The primary coil reduces the circuit it is connected to by a certain voltage so the circuit that the secondary coil is hooked into can serve as a voltage source. In other words, the voltage loss in the first circuit leads to a voltage gain in the other. So if Step 3 yields that the AC voltage drop through the primary coil oscillates between +10 volts and -10 volts, and if n1 is 1000 coils and n2 is 500 coils, then the voltage gain in the secondary coil oscillates between V2 = V1_n2/n1 = +10_(500/1000) = +5 volts and -5 volts.