Answer:
The current through [tex]9 \Omega[/tex] is 0.297 A
Solution:
As per the question:
[tex]R_{5} = 5.0 \Omega[/tex]
[tex]R_{9} = 9.0 \Omega[/tex]
[tex]R_{4} = 5.0 \Omega[/tex]
V = 6.0 V
Now, from the given circuit:
[tex]R_{5}[/tex] and [tex]R_{9}[/tex] are in parallel
Thus
[tex]\frac{1}{R_{eq}} = \frac{1}{R_{5}} + \frac{1}{R_{9}}[/tex]
[tex]R_{eq} = \frac{R_{5}R_{9}}{R_{5} + R_{9}}[/tex]
[tex]R_{eq} = \frac{5.0\times 9.0}{5.0 + 9.0} = 3.2143 \Omega[/tex]
Now, the [tex]R_{eq}[/tex] is in series with [tex]R_{4}[/tex]:
[tex]R'_{eq} = R_{eq} + R_{4} = 3.2143 + 4.0 = 7.2413 \Omega[/tex]
Now, to calculate the current through [tex]R_{9}[/tex]:
[tex]V = I\times R'_{eq}[/tex]
[tex]I = {6}{7.2143} = 0.8317 A[/tex]
where
I = circuit current
Now,
Voltage across [tex]R_{eq}[/tex], V':
[tex]V' = I\times R_{eq}[/tex]
[tex]V' = 0.8317\times 3.2143 = 2.6734 V[/tex]
Now, current through [tex]R_{9}[/tex], I' :
[tex]I' = \frac{V'}{R_{9}}[/tex]
[tex]I' = \frac{2.6734}{9.0} = 0.297 A[/tex]
