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a. Immediately after the switch is closed, are either or both bulbs glowing? Explain. They will both glow as some current passes through them as the capacitor is charging. b. If both bulbs are glowing, which is brighter? Or are they equally bright? Explain. They are both equally bright, because an equal and opposite charge is flowing on to ...

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You're on the right track. You're probably familiar with how the current decreases exponentially after closing the switch. $$I(t)=I_0 e^{-t/\tau}$$ Where $\tau$ is the time constant of the circuit given by $\tau = RC$, and $R$ is the total resistance of the bulbs. So when the switch is closed, current will be a maximum, and the bulbs brightest. As time ...

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You can solve this problem by using Kirchhoff's two laws. Kirchooff's current law tells you that since this is a series circuit the current in each part of the circuit will be the same all the time. So whatever happens to bulb $A$ will also happen to bulb $B$ as they both have the same current flowing through them. Using Kirchhoff's voltage law you have ...

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You're not the first, nor the last, to find the phrase "power flow" somehow wrong. For example, from W J Beaty's article on electrical misconceptions: ELECTRIC POWER FLOWS FROM GENERATOR TO CONSUMER? Wrong. Electric power cannot be made to flow. Power is defined as "flow of energy." Saying that power "flows" is silly. It's as silly as saying that ...

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