Four loads are connected in series across 120 volts and a 3-ampere current exists. One load fail

Four loads are connected in series across 120 volts and a...

Key Concepts

Series Circuit

In a series circuit, all components are connected end?to?end so that the same current flows through each element. Any change in one element, such as an abnormal condition, affects the voltage distribution across the entire circuit.

Voltage Division

Voltage division in a series circuit means that the total voltage supplied is proportional to the resistance of each component. When the resistances are evenly matched or modified by a fault, the voltage drops can change, providing clues about which element might be malfunctioning.

Short Circuit Fault in a Series Chain

A shorted load in a series circuit presents very low or nearly zero resistance, causing it to have a negligible voltage drop. This results in the entire applied voltage being divided among the remaining loads, which may explain why the functional loads each show an increased voltage drop compared to their normal design conditions.

Transcript

00:01 Here in this case the total voltage applied is of 120 volt and because of which the current flows through r1, r2 and r3.

00:10 Voltage across r1 is denoted as vr1, voltage across r2 vr2 and voltage across r3 as vr3.

00:19 Now let's start with step a in which we need to find total power delivered to the three bulbs.

00:29 So i'll write it down total power delivered to three bulbs.

00:41 This is what we need to find out.

00:44 So in this case the first step would be we need to find a resistance of each bulb.

00:50 So now across each bulb, i'll write bulb wattage is equal to 60 watt and bulb voltage is given as 120 volt which is standard for all three of them so therefore first p is equal to as for the formula v square upon r there is a resistance so therefore in order to calculate r what we need to do is v squared divided by p in this case value of v is given as 120 so 120 volt the whole square upon power is given as 60 watt.

01:44 So answer that we get in this case is r is equal to 240 ooms.

01:58 Since all three light bulbs are having the same resistance, since all three light bulbs are having the same resistance.

02:06 Therefore, what we can say, r1 is equal to r2 is equal to r3 is equal to, that would be, 240 ooms, which is again the value.

02:23 Of r that we have so now let's find the equivalent resistance first equivalent resistance resistance of the combination say be are effective we take this notation therefore r -effective is equal to now r2 and r3 they are in parallel and they are in series with r1 so what we can write here that r2 parallel to r3 plus r1 which is equal to 240 oms parallel to again the same that is 240 oms plus 240 oms now how to calculate the parallel combination that is multiplication of the two in the numerator and addition of the two in the denominator so it will go like that so therefore ultimately the value of r effective that we get is 120 plus 240 oms which is equal to 360 oms this is the are effective of the entire combination that we have now we'll mohad next thus we can write thus let total power delivered to the three bulbs be say p t okay that is a total power so therefore p t is equal to same formula we square upon r but here it will be r effective because it is with respect to entire circuit so which is equal to 120 volt square and the value of r effective is 3 60 oms so therefore total power delivered to the three bulbs is of 40 watts this is the main answer for the step a that we have got now now let's start with the step b.

05:12 We'll write it here.

05:16 We need to find a potential difference across each bulb.

05:20 So i'll just write down what we need to find is potential difference across each bulb...

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