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# Kirchoff 's Laws for AP Physics B & C (page 2)

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By McGraw-Hill Professional
Updated on Feb 12, 2011

Now let's trace the circuit, starting at the top left corner and working clockwise:

• The 170 Ω resistor contributes a term of –(170 Ω) I.
• The 1.5 V battery contributes the term of –1.5 volts.
• The 100 Ω resistor contributes a term of –(100 Ω) I.
• The 200 Ω resistor contributes a term of –(200 Ω) I.
• The 9 V battery contributes the term of +9 volts.

Combine all the individual terms, and set the result equal to zero. The units of each term are volts, but units are left off below for algebraic clarity:

0 = (–170)I + (–1.5) + (–100)I + (–200)I + (+9).

By solving for I, the current in the circuit is found to be 0.016 A; that is, 16 milliamps, a typical laboratory current.

The problem is not yet completely solved, though—16 milliamps go through the 100 Ω and 200 Ω resistors, but what about the 300 Ω and 400 Ω resistors? We can find that the voltage across the 170 Ω equivalent resistance is (0.016 A)(170 Ω) = 2.7 V. Because the voltage across parallel resistors is the same for each, the current through each is just 2.7 V divided by the resistance of the actual resistor: 2.7 V/300 Ω = 9 mA, and 2.7 V/400 Ω = 7 mA. Problem solved!

Oh, and you might notice that the 9 mA and 7 mA through each of the parallel branches adds to the total of 16 mA—as required by Kirchoff 's junction rule.

Practice problems for these concepts can be found at:

Circuits Practice Problems for AP Physics B & C

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