Question

A 16 V battery with zero internal resistance is connected to a series combination of a $2\Omega$ resistor that obeys Ohm's law and a thermistor that obeys the current -voltage relation $V=\alpha I+\beta I^2$ with $\alpha =2$ and $\beta =6$. Current through the $2\Omega$ resistor is (in Amperes) (Write up to two digits after the decimal point.)

Answer 1

$16-(\alpha I+\beta I^2)=2I$
$\beta I^2+(2+\alpha )I-16=0\Rightarrow 6I^2+4I-16=0$
$\Rightarrow I=\frac{-4+\sqrt{4^2+4\times 6\times 16}}{2\times 6}=\frac{-4+20}{12}=1.33A$