Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in series with an ideal 12-V battery and a 2.0-Ω resistor, what is the current through the 10.0-Ω resistor

Current across 10ohm, I =0.59 A

Explanation:

Total Resistance n parallel s given as

 1/Rt = 1/R1 + 1/R2 + 1/R3

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

1/Rt = 0.51666

Rt = 1/0.51666

Rt = 1.935 ohms

 which is now connected In series with a ohm resstor

Rnet  in series = R1+ R2

Rnet = 1.935 + 2.0 = 3.935 ohms

Since a voltage of 12 V s connected with the resistors in series, we have that the net current flowing across the 2ohm and 1.935ohm will be

Current, I = Voltage, V/ Resistance, R net

I=  I2 / 3.935= 3.049 A

the Voltage across the 1.935 ohm resistance  

V= 1R

3.049  X  1.935  = 5.899 Volts

 remember that  1.935 ohm s the resultant resistance connected in parallel, therefore The voltage across the 1.935 ohm is same for the 10 ohm, 4ohm and 6ohm resistances

Therefore, the current passing through the 10ohm resistance will be

Current across 10ohm, I   = V/R = 5.899/10 = 0.5899 Ampere  ≈0.59 A

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