i am allowed to use Kirchoff\'s voltage & current laws, reducing resistors in se

Question

i am allowed to use Kirchoff's voltage & current laws, reducing resistors in series and parallel, voltage and current division, source transformation, and ohm's law methods on this

i know i need to find i2 first so i can solve the second circuit as that voltage is in terms of that current, but i'm not quite sure how to do current division if needed. i think the 300 and 60 ohm resistors can also be simplified as they're in parallel, perhaps the same with the 200 and 500 ohm on the other loop

Explanation / Answer

i2 = 50 * 500/(500+200) = 35.71 mA now 300 and 60 are parallel so R = 300 *60/(300+60) = 50 so applying KVL in the loop -35i2 = 60 I s0 -1.25 = 60I => I = -20.8 mA => V1 = 10*(-20.8) x 10^-3 = -0.208 V

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