The peak current to and from a capacitor is 10 mA. What is the...

Question

The peak current to and from a capacitor is 10 mA. What is the peak current if the emf frequency is doubled?

Accepted Answer

Welcome back. Everyone in this problem. A signal is passed through a tuning capacitor in a radio transmission system. The capacitor handles a peak current of 6 million at a particular broadcast frequency. What would be the peak current if the broadcast frequency tripled, keeping the capacitance fixed for our answer choices. A is 25 millions B is 28 million sc is 10 million pires and D is 18 million pires. Now, what are we trying to figure out here? Well, we want to find the peak current if the broadcast frequency tripled. So we basically, we're trying to understand the relationship between the current and the frequency. Once we understand that relationship, then we should be able to figure out the peak current or what would happen to the peak current if we change our broadcast frequency. So first, let's think about it. What do we know about the peak current in a capacitor circuit? Well, recall that we know that the current as flowing through a capacitor equals the potential difference across the capacitor divided by the reactants of the capacitor. And we know that the instantaneous current in a capacitor for an AC circuit can be calculated using Ohm's Law. And that's really what's going on here. It's just that here or reactant it represents is the same or would be the same thing as the impedance Z. OK. Now, what do we know about reactants? Well, we know that our reactant is inversely proportional to our angular frequency and our capacitance. So, since we know that here, we can substitute it into our formula for the current flowing through the capacitor. Because that note tells us that that current will be equal to our voltage divided by our reactants, which is one divided by omega C. And if we simplify that we can rewrite it as a voltage multiplied by omega multiplied by C. So that tells us our current equals our voltage multiplied by our angular frequency multiplied by our capacitance. Now, this is a very important result because this equation tells us that the current to the capacitor is proportional to the capacitance proportional to the frequency of the AC source and to the voltage across the capacitor. So if this is happening for our current, which is usually used for the root mean square current, then at the peak current, OK, add the peak current that tells us, then if we denote this as our peak current, our peak current would be equal to our voltage, which this would still represent our peak voltage multiplied by the broadcast frequency at the peak current multiplied by the capacitance of our circuit. So what does this mean then? Well, what do we know about our peak current? Here? We have our relationship, our peak current is directly proportional to our frequency. So that tells us then that when our broadcast frequency has tripled. In other words, when our broadcast frequency at the peak current has tripled is equal to three multiplied by our angular frequency, then if we substitute that into our formula, we'll find that at the peak current, our formula will be our voltage multiplied by three multiplied by omega multiplied by our capacitance. Now, if we were to rewrite that we could rewrite that as three multiplied by our voltage multiplied by our angular frequency multiplied by capacitance. But what do we already know? We know that voltage multiplied by angular frequency by our capacitance is the same thing as our current. I see the current flowing through the capacitor. So in other words, our peak current equals three multiplied by the current that's flowing through the capacitor, that's our relationship. So when our broadcast frequencies tripled, our peak current also triples. In other words, the current triples when the frequency triples. So to find a new current, when the frequency triples triples, we just need to substitute the value of our current into the equation based on what we knew before we know that our current is 6 million empires at our broadcast frequency. So at the peak current, therefore, our peak current will be equal to three multiplied by 6 million amps, which would be equal to 18 miers. So when the broadcast frequency triples, the peak current would be 18 million. If we look back on our answer choices, that would have been answer choice. D thanks a lot for watching everyone. I hope this video helped.

The peak current to and from a capacitor is 10 mA. What is the peak current if the emf frequency is doubled?

Key Concepts

Capacitive Reactance

Ohm's Law in AC Circuits

Frequency and Current Relationship

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