What is the smallest capacitor needed in a computer power supply in order to supply at least 0.2 A of current for 0.35 s after the power goes off? (DATA: Assume that the resistance of all the components of the computer mother board is 5 Ω and it uses a 5 V supply.)
To determine the smallest capacitor needed, we can use the equation for the discharge of a capacitor:
Q = C * V
Where:
Q is the charge stored in the capacitor
C is the capacitance of the capacitor
V is the voltage across the capacitor
In this case, we want to supply at least 0.2 A of current (I) for 0.35 s (t), and we know that the resistance (R) is 5 Ω. Using Ohm's Law (V = I * R), we can calculate the voltage drop (ΔV) across the components as:
ΔV = I * R
= 0.2 A * 5 Ω
= 1 V
Since the computer motherboard uses a 5 V supply, the initial voltage across the capacitor will be 5 V. Therefore, the voltage across the capacitor after 0.35 s will be:
V = 5 V - 1 V
= 4 V
Now, we can calculate the charge (Q) required by multiplying the current (I) and the time (t):
Q = I * t
= 0.2 A * 0.35 s
= 0.07 C
Finally, we can rearrange the equation Q = C * V to solve for the capacitance (C):
C = Q / V
= 0.07 C / 4 V
= 0.0175 F
Therefore, the smallest capacitor needed in this case would be 0.0175 Farads (or 17.5 millifarads).