Both parts please Bookmarks Poople Window Help 86% ir. Tue 826 AM a The Label On

Question

Both parts please

Bookmarks Poople Window Help 86% ir. Tue 826 AM a The Label On A PXV@Protîrn 5: in The ×VG(3%) Prelem 10 × :0 Aesistors r, Para ×YgGroupMe tructure.com/courses/1118412 8870?module itemjd-11644577 ll 2018)> Assignments > HW04 (6%) Problem 19: In he figure, these three resistors are connected toavoltage source so that R2-425 and R-95 are in parallel with one another and that combination is in series with R 2.52 R. 2.0V R2 Otheexpertta.com 50% Part (a) Calculate the power beng dissipated by the third resistor P-in watts. > 50% Part(b, Find de loal powersupplied bythe scarce, in waes. Grae Smary MacBook Pro 0

Explanation / Answer

R2 and R3 are in parallel so their equivalent

R = R2R3 / (R2 + R3)

= 4.75*9.5 / ( 4.75+9.5)

= 3.167 ohm

R and R1 are in series

R(equi) = R + R1 = 3.167 + 2.5 = 5.67 ohm

I = V/R = 12 / 5.67 = 2.11 A

Current through R3 (I3) = 2.11*4.75 / (4.75+9.5)

= 0.7 A

a) power through R3 = 0.7^2 * 9.5 = 4.88 W

b) power through the battery = IV = 2.11*12 = 25.32 W

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