If the power source is 9.0 volts, the current will be 0.27 amps. The correct answer between all the choices given is the third choice or letter C. I am hoping that this answer has satisfied your query and it will be able to help you, and if you would like, feel free to ask another question.
Answer : Current, I = 0.27 amps
Explanation :
Given that,
Two resistors R₁ = 34.0 ohms and R₂ = 41.0 ohms in parallel and another resistor R₃ = 15.0 ohms in series.
The voltage across them is 9 volts.
Taking the equivalent of R₁ and R₂ :
[tex]\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}[/tex]
[tex]\dfrac{1}{R_{eq}}=\dfrac{1}{34}+\dfrac{1}{41}[/tex]
[tex]R_{eq}=18.58\ \Omega[/tex]
Now, Taking series combination of [tex]R_{eq}[/tex] and R₃
So, [tex]R'_{f}=33.58\ \Omega[/tex]
Using Ohm/s law :
[tex]V=IR[/tex]
[tex]I=\dfrac{V}{R}[/tex]
[tex]I=\dfrac{9\ V}{33.58\ \Omega}[/tex]
I = 0.268 A
or
I = 0.27 amps
So, the correct option is (C) " 0.27 amps "
