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 :
\(\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\)
\(\dfrac{1}{R_{eq}}=\dfrac{1}{34}+\dfrac{1}{41}\)
\(R_{eq}=18.58\ \Omega\)
Now, Taking series combination of \(R_{eq}\) and R₃
So, \(R'_{f}=33.58\ \Omega\)
Using Ohm/s law :
\(V=IR\)
\(I=\dfrac{V}{R}\)
\(I=\dfrac{9\ V}{33.58\ \Omega}\)
I = 0.268 A
or
I = 0.27 amps
So, the correct option is (C) " 0.27 amps "
