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A battery of \(9\ V\) is connected in series with resistors of \(0.2\ Ω\), \(0.3\ Ω\), \(0.4\ Ω\), \(0.5\ Ω\) and \(12\ Ω\), respectively in a circuit. How much current would flow through the \(12\ Ω\) resistor?
 
[Note: Enter the answer up to three decimal places.]
 
The equivalent resistance is given as
 
R=ii+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 \(12\ Ω\) resistor is  \(A\).