What is the voltage across the 6 μf capacitor, Vab = 20V?
Hmmm. I'd kinda like to know a little more about the circuit ...
To determine the voltage across the 6 μF capacitor, we need to know the charge stored on the capacitor. The relationship between the voltage across a capacitor and the charge stored on it is given by the formula:
Q = C * V
where Q is the charge, C is the capacitance, and V is the voltage. Rearranging the formula, we get:
V = Q / C
Now, we know the value of the voltage (V = 20V) and the capacitance (C = 6 μF). Plugging these values into the formula, we can solve for the charge (Q):
Q = C * V
Q = 6 μF * 20V
To perform this calculation, we need to convert the capacitance from microfarads (μF) to farads (F). Since 1 microfarad is equal to 1 × 10^-6 farads, we can convert the capacitance as follows:
C = 6 μF = 6 × 10^-6 F
Now we can substitute the values of C and V into the equation to calculate the charge:
Q = (6 × 10^-6 F) * (20V)
Performing the multiplication, we find:
Q = 0.00012 C
Therefore, the charge stored on the 6 μF capacitor is 0.00012 Coulombs (C).