Three resistors of 4.0, 6.0, and 10.0 Ω are connected in parallel. If the combination is connected in series with a 12.0-V battery and a 2.0-Ω resistor, what is the current through the 10.0-Ω resistor? Group of answer choices

0.536 A

Explanation:

From the question,

The combined resistance of the parallel connection is given as

1/Rt = 1/4+1/6+1/10

Rt = 120/15

Rt = 8 ohms.

Since the the combination is connected in series with a 10 ohms resistor,

Rt' = 8+10 = 18 ohms.

Current flowing through the series circuit

I = V/Rt'

V = 12 V, Rt' = 18 ohms

I = 12/18

I = 0.67 A.

Therefore voltage across the combined resistor

V' = 8(0.67)

v' = 5.36 V.

Since the resistors a connected in parallel, the same voltage drop across them.

Therefore, current through the 10 ohms

Ix = 5.36/10

Ix = 0.536 A

total distance/total time

(85 + 82)/(5+3)

167/8

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