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Three resistors of 12, 12, and 6.0 Ω are connected in parallel. A 12-V battery is connected to the combination. What is the current through the 6.0-Ω resistor?

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Answer:

2 A of electric current is flowing through 6 ohms resistor.

Step-by-step explanation:

Given that,

Resistor 1,
R_1=12\ \Omega

Resistor 2,
R_2=12\ \Omega

Resistor 3,
R_3=6\ \Omega

Voltage of the circuit, V = 12 volts

The equivalent resistance in case of parallel combination of resistors is given by :


(1)/(R)=(1)/(R_1)+(1)/(R_2)+(1)/(R_3)\\\\(1)/(R)=(1)/(12)+(1)/(12)+(1)/(6)\\\\R=3\ \Omega

In parallel combination, current divides in each resistors. Let I is the current through the 6.0 Ω resistor. It can be given by using Ohm's law as :


V=IR\\\\I=(V)/(R_3)\\\\I=(12\ V)/(6\ \Omega)\\\\I=2\ A

So, 2 A of electric current is flowing through 6 ohms resistor. Hence, this is the required solution.

User Harminder Singh
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