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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 ? |
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Answer» SOLUTION :Here `C_1 = C_2 = C_3 = 9 pF` (a)In SERIES combination the resultant capacitance `1/C= 1/C_1+ 1/C_2 + 1/C_3= 1/9 + 1/9 + 1/9 = 1/3 rArrC = 3pF= 3xx 10^(-12) F`. (b) As total POTENTIAL difference V = 120 V `V_1+ V_2 + V_3 = 120V` As all the 3 CAPACITORS are identical, hence `V_1 = V_2= V_3= V/3 = 120/3= 40V` |
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