Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Given: The capacitance of each capacitor is 9 pF .
(a)
Total capacitance of the capacitor when connected in series is given as,
1 C t = 1 C 1 + 1 C 2 + 1 C 3
where, the total capacitance is C t , C 1 , C 2 and C 3 are the capacitance of each capacitor.
By substituting the given values in above equation, we get
1 C t = 1 9 + 1 9 + 1 9 1 C t = 3 9 C t = 3 pF
Thus, the total capacitance of the capacitor when connected in series is 3 pF .
(b)
The combination is connected by 120 V supply.
For equal capacitance of the capacitor, the potential difference of each capacitor is given as,
V ' = V n
where, potential difference of each capacitor is V ' , voltage supply is V and number of equal capacitor in series is n .
By substituting the given values in the above equation, we get
V ' = 120 3 = 40 V
Thus, the potential difference of each capacitor is 40 V .
Given: The capacitance of each capacitor is
(a)
Total capacitance of the capacitor when connected in series is given as,
where, the total capacitance is
By substituting the given values in above equation, we get
Thus, the total capacitance of the capacitor when connected in series is
(b)
The combination is connected by
For equal capacitance of the capacitor, the potential difference of each capacitor is given as,
where, potential difference of each capacitor is
By substituting the given values in the above equation, we get
Thus, the potential difference of each capacitor is
