Answer:
The answer is below
Explanation:
The complete question is in the iimage attached.
a) We can see that R1 is in series with the parallel combination of R2 and R3. Hence the equivalent resistance is given by:
[tex]R_{eq}=R_1+(R_2//R_3)\\\\R_{eq}=R_1+\frac{R_2R_3}{R_2+R_3}\\ \\R_{eq}=6+\frac{6*6}{6+6}\\\\R_{eq}=6+3\\\\R_{eq}=9\ \Omega[/tex]
b)
The current in the circuit (I) is given as:
I = emf / Req
I = 18 / 9
I = 2A
The current through resistor R1 = I1 = I = 2A
Since R2 and R3 are parallel to each other their respective currents I2 and I3 is given as:
[tex]I_2=I*\frac{R_3}{R_2+R_3}=2*\frac{6}{6+6}=1\ A \\\\I_3=I*\frac{R_2}{R_2+R_3}=2*\frac{6}{6+6}=1\ A[/tex]
