Answer:
2 A of electric current is flowing through 6 ohms resistor.
Explanation:
Given that,
Resistor 1, [tex]R_1=12\ \Omega[/tex]
Resistor 2, [tex]R_2=12\ \Omega[/tex]
Resistor 3, [tex]R_3=6\ \Omega[/tex]
Voltage of the circuit, V = 12 volts
The equivalent resistance in case of parallel combination of resistors is given by :
[tex]\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\\\\\dfrac{1}{R}=\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{6}\\\\R=3\ \Omega[/tex]
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 :
[tex]V=IR\\\\I=\dfrac{V}{R_3}\\\\I=\dfrac{12\ V}{6\ \Omega}\\\\I=2\ A[/tex]
So, 2 A of electric current is flowing through 6 ohms resistor. Hence, this is the required solution.
