Two capacitors with capacitances of 3.25 and 5.00 μF, respectively, are connected in parallel. The system is connected to a 55-V
1 answer:
The charge accumulated in 3.25 μF capacitor is 178.75 μC.
Answer:
Explanation:
In parallel connection, the voltage drop across any passive devices like capacitor or resistor will be constant. So the current flow will be varying in case of parallel connections of capacitors or resistors.
As the capacitance is the measure of amount of charge or current generated for a given amount of voltage, it is directly proportional to the charge or current and inversely proportional to voltage.
C = Q/V
Here the charge accumulated in a capacitor of capacitance 3.25 microfarad need to be determined which is in parallel connection with another capacitor. So the voltage through both the capacitor will be equal to the voltage of the battery which is stated as 55 V.
3.25× = Q/55
Q = 3.25 * 55 = 178.75 μC
So the charge accumulated in 3.25 μF capacitor is 178.75 μC.
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