Cividual 1. A charge of \( 1,67 \mathrm{C} \) flows through the filament of a particular light bulb in \( 2,00 \mathrm{~s} \). (a) Calculate the current flowing through the light bulb (b) Calculate the number of electrons that pass through the filament every second. A current of 5 A flows through a wire for 4,0 minutes. How much charge passe through any given cross-section of the wire?

The current flowing through the light bulb can be calculated using the formula \( I = \frac{Q}{t} \), where \( Q \) is the charge in coulombs and \( t \) is the time in seconds. Given \( Q = 1.67 \, \text{C} \) and \( t = 2.00 \, \text{s} \), the current \( I \) is \( \frac{1.67}{2.00} = 0.835 \, \text{A} \). To find the number of electrons passing through the filament every second, use the charge of an electron \( e = 1.6 \times 10^{-19} \, \text{C} \). The total charge per second (current) is \( 0.835 \, \text{C} \). Therefore, the number of electrons is \( \frac{0.835}{1.6 \times 10^{-19}} \approx 5.22 \times 10^{18} \) electrons. Now, for the wire carrying a current of 5 A for 4 minutes, we first convert the time into seconds: \( 4 \, \text{min} = 240 \, \text{s} \). The total charge \( Q \) can be calculated using \( Q = I \cdot t = 5 \, \text{A} \times 240 \, \text{s} = 1200 \, \text{C} \). So, 1200 coulombs of charge have passed through a cross-section of the wire during this time.