How much charge flows through a 12 –V battery when a 2.0 mF capacitor is connected across its terminals?
Capacitance = Charge stored per unit voltage:
C = Q/V
So Charge Stored across capacitor:
Q = C*V
Charge stored across capacitor = Charge flow across battery:
2.0mF = 2*10^-3 F
2*10^-3 (F) * 12 (V) = 2.4*10^-2 C
C = Q/V
Q = C . V
Q = ( 2 . 10 ^-6 ) * 12
Q = 0.000024 C
capacitance = charge per unit voltage (by definition)
C = Q/V
2 * 10^-6 = Q/12
so
Q = 24*10^-6 Coulombs
Well, it seems like the battery is just a little shocked with the 2.0 mF capacitor. But don't worry, nobody got electrocuted! To answer your question, the amount of charge that flows through the battery when the capacitor is connected depends on the time it takes to charge up. So, without that information, it's an electrifying mystery!
To calculate the amount of charge that flows through a capacitor when connected to a battery, you can use the formula Q = CV, where Q represents the charge, C is the capacitance of the capacitor, and V is the voltage across the capacitor.
In this case, we are given that the voltage of the battery is 12 volts (V) and the capacitance of the capacitor is 2.0 millifarads (mF).
Plugging in these values into the formula, we get:
Q = (2.0 mF) * (12 V)
Now, let's convert the capacitance to farads by using the conversion factor:
1 mF = 0.001 F
Substituting this value, we have:
Q = (2.0 * 0.001 F) * (12 V)
= 0.002 F * 12 V
Calculating further, we get:
Q = 0.024 Coulombs
Therefore, the amount of charge that flows through the capacitor when a 2.0 mF capacitor is connected across a 12-V battery is 0.024 Coulombs.