Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. If the two light bulbs are connected in series across a 120-V line, find (b) the power dissipated in each bulb.

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Hey everyone today we're dealing with a problem about circuits. So we're being told that a circuit has two heating coils that are connected in series. Now the coils are one in our to have resistance is of and 550 homes respectively. We also know that the combination is connected to a 110 volt power source. So with this information we're being asked to calculate the rate of power dissipation in each coil so each coils separately. So before we go ahead and start working on this, I'd like to um point out a few formulas to remember. We can recall that for power in watts can be equal to the potential difference, multiplied by the current in amperes. This can also be equivalent to I squared R. What are as resistance in homes which is also equivalent to v squared divided by R. We can also utilize owns Law rooms. Law to find out exactly what the current is in our series as well. The sequel to I. R. So with this in mind, let's go ahead and draw what this scenario looks like. We have a voltage source voltage source of 110V. We have a one resistor and the second resistor Where this is R one and R two respectively. So in a series circuit the potential difference applied across each resistor is less than the potential difference supplied by the power source. So the voltages across R one and R two will be less than 110 volts. And this makes sense because we're being passed through a resistance. However, in series current is the same through both. Let's write that out current. I let's write that. Normally the current I his same Through R. one and R two. So if we can find out the equivalent resistance in the circuit, then we can go ahead and determine what the current is using Homes Law. So In a series circuit, equivalent resistance or total resistance is equal to the sum of each resistance or each resistor. That is in that series connection. So this will be 350 plus 550. Giving us a final answer of 900 homes. 900 homes plugging this back into Homes Law and rearranging for I we get that i is equal to V by R. which is equal to 110V divided by 900 homes, giving a final current of 0.122 and pierce. So with this and the equivalent resistance already here we can go ahead and actually solve for the power dissipation across each coil. So let's write the sound power for the first coil is equal to I Squared into R one, which is equal to 0.122 Amperes squared Multiplied by 350, which gives us an answer Of 5.21 watts. Similarly for the power through the second resistor, it's done very similarly with the Current multiplied by R two instead, which gives us 0.122 amperes squared Multiplied by which gives a final answer of 8.19 watts, so the power dissipated through resistor one or coil one is 5.21 watts, and through coil two is 8.19 watts or answer choice C. I hope this helps, and I look forward to seeing you all in the next one.

Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. If the two light bulbs are connected in series across a 120-V line, find (b) the power dissipated in each bulb.

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Key Concepts

Ohm's Law

Power in Electrical Circuits

Series and Parallel Circuits