2. Choose the correct answer If the duration of the pulse front is 1mu s and the maximum voltage value is 1 mV, then the steepness of pulse front is equal to __ 0.8cdot 10^-3s/V 0.8cdot 10^-3V/s 0.8cdot 10^3s/V 0.8cdot 10^3V/s

The correct answer is:<br /><br />$0.8\cdot 10^{3}V/s$<br /><br />Explanation:<br />The steepness of a pulse front is defined as the rate of change of voltage with respect to time. It is calculated by dividing the maximum voltage value by the duration of the pulse front.<br /><br />Given:<br />- Duration of the pulse front = $1\mu s$ = $1 \times 10^{-6}$ seconds<br />- Maximum voltage value = 1 mV = $1 \times 10^{-3}$ volts<br /><br />The steepness of the pulse front is:<br /><br />$\text{Steepness} = \frac{\text{Maximum voltage value}}{\text{Duration of pulse front}}$<br /><br />$\text{Steepness} = \frac{1 \times 10^{-3} \text{ volts}}{1 \times 10^{-6} \text{ seconds}}$<br /><br />$\text{Steepness} = 1 \times 10^{3} \text{ volts/second}$<br /><br />Therefore, the correct answer is $0.8\cdot 10^{3}V/s$.