This voltage can be viewed as the sum of two voltages, V 1 a + V 2 a, where V 1 a becomes +50 volts at t 0 and remains there indefinitely, and V 2 a becomes 50 volts. Superposition: Two Loop Problem To apply the superposition theorem to calculate the current through resistor R 1 in the two loop circuit shown, the individual current supplied by each battery is calculated with the other battery replaced by a short circuit. The voltage at a starts at zero, goes to +50 volts at t 0, then returns to zero at t +0.001 second. No matter what path you take from node X to node Y, you must get the same potential difference between X and Y.ĭo the same for the current source and add the two?įor a basic rundown of how to solve circuits by superposition, see here. The superposition principle (see above) is used to solve the problem. It's simply the consequence of being connected in parallel and Kirchoff's voltage law. It has nothing to do with the current division theorem. Since \$\sin(\omega_1 t)\$ and \$\sin(\omega_2 t)\$ are orthogonal functions when \$\omega_1\ne\omega_2\$, there's no way to simplify \$\sin(\omega_1 t) + \sin(\omega_2 t)\$ it's already the simplest form you'll be able to write.ĭue to the current division theorem, the voltage across the capacitor is also the same voltage across the inductor? How does differing angular frequencies change the summation of the two sources?
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