sl Little Wing 403 Aumac x Hats outdoor Militias x LON-CAPI, current in ea x M V

Question

sl Little Wing 403 Aumac x Hats outdoor Militias x LON-CAPI, current in ea x M VT --o 3566 Step-by- x Ce Secure l https capag phy ohio. e res/wiley/hrw/9e/chapter2/2/-014 problem?symb uploaded%2fohiou%2f8x13014f1bb4a580boucapa2%2fdefault 145575858012648851456536 /2 lb2esequence 19 wi ua welcome, Timothy Os Mail Vilson, Timoth e LON-CAPM The Leam c Chegg study I Guided b CE2010. mothy i Current in each resistor Due in 3 hours, 42 minutes As shown in the figure, R1 85 R2 R3 40 s?, R4 70 2, and the ideal battery has e E 24 v, what is the equivalent resistance? 1.01x102 ohms You are correct Your recept no s 131 What is the current in resistance 1? 2.39x10 A You are correct Your receipt no s 151. 293 What is the current in resistance 2? 29A Submit Answer incorrect. es 2/10 previous Tries What is the current in resistance 37 0929mA Submit Ariswer Incorrect. es 1/10 DreviousInies What esistance 4 Sub Answer es 0/10 Ask me any 3/13/20

Explanation / Answer

Req = 1.01 x 10^2 Ohm

I = V/R = 24/1.01 x 10^2 = 2.38 x 10^-1 A

Current through R1 will beequal to the battery current. So drop across R1 will be:

V1 = 2.38 x 10^-1 x 85 = 20.23 Volts

Voltage across parallel cobination is:

Vp = 24 - 20.23 = 3.77 V

Since voltage is same in parallel,

V2 = 3.77 V, I2 will be:

I2 = V2/R2 = 3.77/40 = 0.0943 = 9.43 x 10^-2 A

Hence, I2 = 9.43 x 10^-2 A

I3 = V3/R3 = 3.77/40 = 9.43 x 10^-2 A

Hence, I3 = 9.43 x 10^-2 A

I4 = V4/R4

I4 = 3.77/70 = 5.39 x 10^-2 A

Hence, I4 = 5.39 x 10^-2 A

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