The value of a series resistor required to limit the current through an electric bulb to 40 mA with a forward voltage drop of 4V when connected to 16 V supply is ________.

Correct option is D

$$\text{Given:} \\[4pt]\bullet \; \text{Supply Voltage, } V_T = 16\,\text{V} \\[4pt]\bullet \; \text{Forward Voltage of Bulb, } V_{\text{Bulb}} = 4\,\text{V} \\[4pt]\bullet \; \text{Current, } I = 40\,\text{mA} = 0.04\,\text{A} \\[4pt]\bullet \; \text{Voltage drop across the series resistor, } V_R = V_T - V_{\text{Bulb}} \\[4pt]\bullet \; \text{The value of the series resistor is given by } R = \frac{V_R}{I} \\[8pt]\text{Calculate } V_R: \\[4pt]V_R = V_T - V_{\text{Bulb}} = 16\,\text{V} - 4\,\text{V} = 12\,\text{V} \\[8pt]\text{Now calculate the value of the resistor } R: \\[4pt]R = \frac{V_R}{I} = \frac{12\,\text{V}}{0.04\,\text{A}} = 300\,\Omega$$​​