Two capacitors, C1 = 5 μF and C2 = 12 μF, are connected in parallel,
and the resulting combination is connected to a 9 V battery. Find (a) the equivalent
capacitance of the combination, (b) the potential difference across each capacitor, and
(c) the charge stored on each capacitor.
very bad extremely bad ,chal nikal
To find the equivalent capacitance of the combination, you need to use the formula for capacitors connected in parallel, which states:
1/Ceq = 1/C1 + 1/C2
where Ceq is the equivalent capacitance. Plugging in the given values, you can calculate:
1/Ceq = 1/5μF + 1/12μF
To add these fractions, you need a common denominator:
1/Ceq = (12/12) * (1/5μF) + (5/5) * (1/12μF)
= 12/60μF + 5/60μF
= 17/60μF
Now, take the reciprocal of both sides to find Ceq:
Ceq = 60μF/17
Therefore, the equivalent capacitance of the combination is approximately 3.53μF (rounded to two decimal places).
To find the potential difference across each capacitor, note that capacitors connected in parallel have the same potential difference. In this case, the potential difference is 9 V, as given by the battery. So each capacitor will have a potential difference of 9 V.
To find the charge stored on each capacitor, you can use the formula:
Q = CV
where Q is the charge, C is the capacitance, and V is the potential difference. Plugging in the values, you have:
For C1:
Q1 = (5μF)(9V)
= 45μC
For C2:
Q2 = (12μF)(9V)
= 108μC
Therefore, the charge stored on C1 is 45μC, and the charge stored on C2 is 108μC.
Don't understand
add capacitance of capacitors in parallel (holds sum of charges at same voltage, which answers second question)
C = charge/unit voltage
so
charge = C V = 9 times C for each