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NOTE Ohm's Law

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<v Narrator>One equation, three questions. Ohm's law relates the three quantities from the first lesson: E equals I times R Voltage equals current times resistance.

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<v Narrator>Rearranged: I equals (E) divided by (R) R equals (E) divided by (I) That is the entire content of the law. The exam difficulty is never the algebra; it is spotting which quantity the question gave you and which it wants.

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<v Narrator>Reading the question. Pool questions in this group are all one shape: two values are stated with their units, and the third is asked for. The units are the tell. Volts and ohms given → they want current, so divide.

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<v Narrator>Volts and amps given → they want resistance, so divide. Amps and ohms given → they want voltage, so multiply.

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<v Narrator>Notice that two of the three cases are division, and the one multiplication is the case where neither given value is in volts. If you remember nothing else: if volts is one of your givens, you divide.

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<v Narrator>Worked examples. What is the resistance of a circuit in which a current of 3 amperes flows through a resistor with 90 volts applied? Volts and amps given, so divide: R = E / I = 90 / 3 = 30 ohms.

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<v Narrator>What is the current through a 100 ohm resistor connected across 200 volts? Volts and ohms given, so divide: I = E / R = 200 / 100 = 2 amperes.

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<v Narrator>What is the voltage across a 2 ohm resistor if a current of 0.5 amperes flows through it? Amps and ohms, so multiply: E = I times R = 0.5 times 2 = 1 volt.

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<v Narrator>Watch the prefixes. The pool likes to combine Ohm's law with the previous lesson. A question that gives you 500 milliamperes and 100 ohms is asking you to convert first: 500 mA = 0.5 A, so E = 0.5 times 100 = 50 volts.

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<v Narrator>Working the arithmetic in base units, amps, volts, ohms, and converting only at the end is the habit that prevents this.

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<v Narrator>The sanity check. Before writing an answer, ask whether its size is plausible: A large resistance with a modest voltage gives a small current. A small resistance with a modest voltage gives a large current.

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<v Narrator>If you calculated 200 amps through a resistor in a handheld radio, you have divided the wrong way round. This check catches essentially every mistake that comes from grabbing the wrong rearrangement under time pressure.

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<v Narrator>Where Ohm's law stops working. Ohm's law describes resistance. It does not, by itself, describe capacitors, inductors, or antennas, whose opposition to AC changes with frequency.

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<v Narrator>For those you replace R with impedance Z and the arithmetic becomes more involved, that is the reactance module, later. It also assumes a fixed temperature.

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<v Narrator>A lamp filament's resistance rises sharply as it heats, which is why a cold bulb draws a large inrush current for the first instant after switch-on. Check yourself. A 12 V supply drives 250 mA through a load.

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<v Narrator>What is the load's resistance? What current flows through a 4700 ohm resistor across 9 V? A resistor drops 3.3 V while carrying 22 mA. What is its value?

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<v Narrator>Convert first: 250 mA = 0.25 A. R = E / I = 12 / 0.25 = 48 ohms. I = E / R = 9 / 4700 = 0.00191 A = 1.9 mA. Small resistance question, small answer, plausible. 22 mA = 0.022 A. R = 3.3 / 0.022 = 150 ohms.
