Answer:
I = 1.09 A
Explanation:
Here, 5 ohm and 12 ohm resistor are connected in parallel. Using the parallel combination of resistors as follows :
[tex]\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{R}=\dfrac{1}{5}+\dfrac{1}{12}\\\\ R=3.52\ \Omega[/tex]
Now R and 2 ohm resistors are in series. Let the equivalent be [tex]R_{eq}[/tex]. So,
[tex]R_{eq}=3.52+2\\\\=5.52\ \Omega[/tex]
Let I be the current. Using Ohm's law,
V = IR
[tex]I=\dfrac{V}{R_{eq}}\\\\I=\dfrac{6}{5.52}\\\\I=1.086\ A[/tex]
or
I = 1.09 A
So, the correct option is (c).
