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MIT 6.002 Circuits and Electronics, Spring 2007
11 Online
12V battery in series with 3 resistros. Given: R1=1 ohm Voltage drop accross R2=4V Power accross R3=12V. What is current in the circuit?
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I^2 - 8 I +12 = 0 is the correct quadratic equation I = - (-8/2) [1 +- sqrt(1 - 12/(8/2)^2)] I = 4 [1 +- sqrt((1- 12/16)] I = 4 [1 +- sqrt{1 - 3/4)] I = 4 [1 +- 1/2] I = 6, 2 The choice of the two answers would be dependent on the required voltage across the 1 ohm resistor which is dependent on R2 and R3
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