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- Thread starter butz3
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cepheid

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First lets look at Kirchoff's Loop Rule, what we usually call Kirchoff's Voltage Law (KVL) in my class. It states that the sum of the potential differences around the loop must be equal to zero. In this case, that means that the voltage across the source is equal to the sum of the potential drops across the four resistors:

[tex] V_s - (V_1 + V_2 + V_3 + V_4) = 0 [/tex]

Let's talk about the current in the circuit. This being a loop (all elements in series), the current must be the same everywhere in the circuit. Detectors at any two different points along the loop should measure the same rate of charge per second flowing past, otherwise charge is piling up somewhere, or charge is not conserved. Since we have a steady current, and charge is conserved, the current is constant and equal to [itex] I [/itex] everywhere.

Ohm's law tells us two things:

1. This total current [itex] I [/itex] in the circuit is equal to the voltage of the source divided by the total (or equivalent) resistance of the circuit:

[tex] I = \frac{V_s}{R_{eq}} [/tex]

[tex] \therefore \ \ V_s = R_{eq}I [/tex]

2. The potential difference (ie voltage drop) across each resistor is given by the current through it times its resistance:

[tex] V_1 = R_{1}I [/tex]

[tex] V_2 = R_{2}I [/tex]

[tex] V_3 = R_{3}I [/tex]

[tex] V_4 = R_{4}I [/tex]

Substitute the results of 1 and 2 (from Ohm's law) into the the very first equation (from KVL):

[tex] V_s - (V_1 + V_2 + V_3 + V_4) = 0 [/tex]

[tex] R_{eq}I - (R_{1}I + R_{2}I + R_{3}I + R_{4}I) = 0 [/tex]

Cancel out I and rearrange terms:

[tex] R_{eq} = R_{1} + R_{2} + R_{3} + R_{4} [/tex]

That's the result we were trying to prove.

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