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OpenStudy (anonymous):

someone check my work. (three-branch parallel network)

OpenStudy (anonymous):

A 2.0-ohm resistor is connected in series with a 20.0-V battery and a three-branch parallel network with branches whose resistances are 8.0 ohms each. Ignoring the battery's internal resistance, what is the current in the battery? Show your work.

OpenStudy (anonymous):

I = V / R total resistance = 2 ohm + the total resistance of the 3 parallel resistances 1/R = 1/R1 + 1/R2 + 1/R3 1/R = 1/8 + 1/8 + 1/8 = 3/8 R = 2.6667 So total Resistance of the circuit = 2.7 + 2.0 = 4.7 I = voltage /Reistance 20/4.7 = 4.25 A

OpenStudy (anonymous):

yes that should be correct

OpenStudy (kropot72):

The answer rounded to 3 decimal places is 4.286 A

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