Answer:
R = 42.67 ohms
Explanation:
It is given that, a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A.
It means that when two batteries are connected in series, then the total voltage is 3.2 volts
Let R is the resistance of the glowing thin-filament bulb. So, using Ohm's law we get :
[tex]V=IR\\\\R=\dfrac{V}{I}[/tex]
So,
[tex]R=\dfrac{3.2}{0.075}\\\\R=42.67\ \Omega[/tex]
So, the resistance of the bulb is 42.67 ohms.
