Four resistors are connected to a battery with a terminal...

Four resistors are connected to a battery with a terminal voltage of 12 V, as shown in the figure below. (Assume R1 = 27.0 ?? and R2 = 71.0 ??.) (a) How would you reduce the circuit to an equivalent single resistor connected to the battery? Use this procedure to find the equivalent resistance of the circuit. (b) Find the current delivered by the battery to this equivalent resistance. (c) Determine the power delivered by the battery. (d) Determine the power delivered to the 50.0-?? resistor.

Transcript

00:01 Hi, here in this given problem first of all there is a mixed grouping of the resistors.

00:07 First of all this r1 that is in series with 50 .0 om and then both of them are in parallel with r2 and then r no sorry this 20 om is in series with these three and then a battery which is always in series providing 12 volt so here the value of r1 that is 27 .0 om value of r2 this is 71 .0 om in the first part of the problem we have to find an equivalent resistance which may be replaced which may replace these four resistances so here first of all r1 and 50 are in series as r1 is in series with this 50 .0 om so the net resistance r s will be given by r1 plus 50 27 .0 plus 50 .0 ome so it comes out to be equal to 77 .0 om now this combined r s that will be in parallel with r2 now this r s is in parallel with r2 so now this r s is in parallel with r2 so now they're combined resistance will be given by r p stands for parallel and that is having a shortcut in numerator their product r s into r2 in denominator their addition rs plus r2 so here it is 77 multiplied by 71 divided by 77 plus 71 so it comes out to be equal to 36 .9 om now this point parallel rp.

02:45 Combinedly it is in series with the very first resistance means that is 20 om.

03:01 So r equivalent equivalent resistance with which this circuit may be replaced, it comes out to be equal to r p plus 20 means that is 56 .9 om.

03:23 Which is the answer for the first part of the problem.

03:29 Then in the second part of the problem, we have to find current delivered by the battery, and that will be using oms law, i is equal to v by r, here r equivalent.

03:59 So this is 12 volt divided by 56 .9 om.

04:05 So finally this current delivered by the battery comes out to be equal to.

04:12 0 .21 ampere answer for the second part of the problem then in the third part of the problem we have to find power delivered by the battery and for this we use the expression i squared total current provided by the battery into our equivalent resistance total resistance so here it will be 0 .21 square multiplied by 56 .9 so this power delivered by the battery comes out to be equal to 2 .53 watt it comes out to be equal to then finally in the fourth and the last part of the problem we have to find power delivered to 50 om here if you look in the the circuit...

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