Question

Two resistors of 5.0 and 9.0 ohms are connected inparallel. A

Answer 1

Answer:

The current through $9 \Omega$ is 0.297 A

Solution:

As per the question:

$R_{5} = 5.0 \Omega$

$R_{9} = 9.0 \Omega$

$R_{4} = 5.0 \Omega$

V = 6.0 V

Now, from the given circuit:

$R_{5}$ and $R_{9}$ are in parallel

Thus

$\frac{1}{R_{eq}} = \frac{1}{R_{5}} + \frac{1}{R_{9}}$

$R_{eq} = \frac{R_{5}R_{9}}{R_{5} + R_{9}}$

$R_{eq} = \frac{5.0\times 9.0}{5.0 + 9.0} = 3.2143 \Omega$

Now, the $R_{eq}$ is in series with $R_{4}$:

$R'_{eq} = R_{eq} + R_{4} = 3.2143 + 4.0 = 7.2413 \Omega$

Now, to calculate the current through $R_{9}$:

$V = I\times R'_{eq}$

$I = {6}{7.2143} = 0.8317 A$

where

I = circuit current

Now,

Voltage across $R_{eq}$, V':

$V' = I\times R_{eq}$

$V' = 0.8317\times 3.2143 = 2.6734 V$

Now, current through $R_{9}$, I' :

$I' = \frac{V'}{R_{9}}$

$I' = \frac{2.6734}{9.0} = 0.297 A$