How to Measure Two Resistors in Parallel
Consider, as an example of resistors in parallel, a setup with a 9-V battery powering a circuit with a single wire leading away from the negative terminal and passing through an initial resistor, R1. The wire continues on before splitting into two wires, each passing through their own individual resistor, R2 and R3. Then the two wires rejoin before connecting again to the battery, this time to its positive terminal. Suppose also that you have an ammeter for measuring the current passing through the three resistors. Call them i1, i2 and i3. Suppose, finally, that R2 and R3 are unknown quantities, which you need to determine.
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Instructions
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Measure the currents going through the three resistors with an ammeter. Suppose, for this example, that you find i1 = 1 amp, i2 = 1/3 amp and i3 = 2/3 amp. Note that i2 + i3 = i1 is required to obey Kirchhoff's junction rule.
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Apply Kirchhoff's loop rule, which says that whichever of the two circuits an electron from the battery travels to get from the negative to the positive terminal, it must experience the same potential drop. Therefore, for battery voltage, V, you have two equations: V = iR1 + i2 x R2 and V = iR1 + i3 x R3. Suppose that you can select R1, because the point of the circuit is to determine R2 and R3. Then you can quickly solve for R2 and R3 from these equations alone.
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Use known R1 to solve for R2 and R3. Continuing with the above example, suppose you switched in R1 of known resistance 1 ohm. Then V = iR1 + i2 x R2 gives R2 = 24 ohms. The equation V = iR1 + i3 x R3 gives 12 ohms.
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Tips & Warnings
Note that the equation of the effective resistance of R2 and R3 didn't need to be invoked to solve the problem. Don't conclude, therefore, that if R1 wasn't known that the effective resistance equation could be invoked to give the necessary third equation to solve three unknowns. Two of these equations would have been redundant, and you'd be stuck with only two equations in three unkowns.
