Three resistors having resistances of R 1 = 1.77 , R 2 = 2.65 and R 3 = 4.92 res

Question

Three resistors having resistances of R1 = 1.77 , R2 = 2.65 and R3 = 4.92 respectively, are connected in series to a 27.7 V battery that has negligible internal resistance

Part A

Find the equivalent resistance of the combination.

Req=?

Part B

Find the current in each resistor.Answer in the order indicated. Separate your answers using commas.

IR1,IR2,IR3 =?

Part C

Find the total current through the battery.

I=?

Part D

Find the voltage across each resistor.

Answer in the order indicated. Separate your answers using commas.

VR1,VR2,VR3=?

Explanation / Answer

Three resistors  are connected in series to a 27.7 V battery that has negligible internal resistance,therefore current through all 3 resistors is same.

part (a) Req = R1 +R2 + R3 = 1.77 + 2.65 + 4.92 = 9.34 ohms

part (b) resistors  are connected in series , ,therefore current through all 3 resistors is same.

using V= iR

27.7V = i(9.34ohm)

i = 2.9657 Amp.

iR1 = 2.9657 A   iR2 = 2.9657 A   iR3 = 2.9657 A

part(c)   total current through the battery = 27.7V / 9.34 ohms = 2.9657 amp.

part(d) voltage across R1 = i*(R1) = 2.9657 * 1.77 = 5.249289 V

voltage across R2 = i*(R2) = 2.9657 * 2.65 = 7.859105 V

voltage across R3 = i*(R3) = 2.9657 * 4.92 = 14.591244 V

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