Find the correct value below:
 
A battery of \(6\ V\) is connected in series with resistors of \(2\ Ω\), \(4\ Ω\), \(8\ Ω\), and \(11\ Ω\), respectively in a circuit. How much current would flow through the \(11\ Ω\) resistor?
 
The equivalent resistance is given as
 
R=ii+ii+ii+ii
 
On substituting the known values, we get \(R\ =\)  \(Ω\)
 
Using Ohm's law, we get
 
I=ii
 
The current flowing across the \(11\ Ω\) resistor is  \(A\).
 
[Note: Enter the answer in two decimal place]