Question

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

4.0 Ohm resistor is then connected in series withthe parallel
combination. A 6.0V battery is then connected tothe series-parallel
combination. What is the current throughthe 9.0 ohm resistor?

Answer 1

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

`(1)/(R_(eq)) = (1)/(R_(5)) + (1)/(R_(9))`

`R_(eq) = (R_(5)R_(9))/(R_(5) + R_(9))`

`R_(eq) = (5.0* 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* R'_(eq)`

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

where

I = circuit current

Now,

Voltage across
`R_(eq)`, V':

`V' = I* R_(eq)`

`V' = 0.8317* 3.2143 = 2.6734 V`

Now, current through
`R_(9)`, I' :

`I' = (V')/(R_(9))`

`I' = (2.6734)/(9.0) = 0.297 A`