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Class 12 Tuition > Learn Exercise 2 > Three capacitors each of capacitance 9 pF are connected...
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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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(a) Capacitance of each of the three capacitors, C = 9 pF Equivalent capacitance (C’) of the combination of the capacitors is given by the relation,1C'=1C+1C+1C⇒1C'=19+19+19=13⇒C'=3 pF Therefore, total capacitance of the combination is 3 pF . (b) Supply voltage, V = 120 V Potential...
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(a) Capacitance of each of the three capacitors, C = 9 pF Equivalent capacitance (C’) of the combination of the capacitors is given by the relation,1C'=1C+1C+1C⇒1C'=19+19+19=13⇒C'=3pF Therefore, total capacitance of the combination is 3pF . (b) Supply voltage, V = 120 V Potential difference (V') across each capacitor is equal to one-third of the supply voltage. Therefore, the potential difference across each capacitor is 40 V. read less
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Engineer and a civil services exam aspirant.

The capacitance of each of three capacitors =9pf(a) equivalent capacitance= C'1/C'= 1/C1 + 1/C2 + 1/C3 = 1/9 + 1/9 + 1/91/C' = 3/9C' = 3pf (b) supply voltage (V)=120Vpotential difference across each capacitor is 1/3rd of supply voltage.i.e. V'= V/3 V' = 120/3 V' = 40 volts V'...
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The capacitance of each of three capacitors =9pf(a) equivalent capacitance= C'1/C'= 1/C1 + 1/C2 + 1/C3 = 1/9 + 1/9 + 1/91/C' = 3/9C' = 3pf (b) supply voltage (V)=120Vpotential difference across each capacitor is 1/3rd of supply voltage.i.e. V'= V/3 V' = 120/3 V' = 40 volts V' = 40 volts read less
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Mechanical engineer with 2 years of teaching experience

The capacitance of each of three capacitors =9pf(a) equivalent capacitance(C') 1/C'= 1/C1 + 1/C2 + 1/C3 = 1/9 + 1/9 + 1/9 1/C' = 3/9 C' = 3pf(b) supply voltage (V)=120Vpotential difference across each capacitor is 1/3rd of supply voltage.i.e. V'= V/3 V' = 120/3 V' = 40 volts V' = 40 vo...
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The capacitance of each of three capacitors =9pf(a) equivalent capacitance(C') 1/C'= 1/C1 + 1/C2 + 1/C3 = 1/9 + 1/9 + 1/9 1/C' = 3/9 C' = 3pf(b) supply voltage (V)=120Vpotential difference across each capacitor is 1/3rd of supply voltage.i.e. V'= V/3 V' = 120/3 V' = 40 volts V' = 40 volt read less
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Given, C1 = C2 = C3 = 9 pF = 9 x 10-12 F V = 120 volt a) Total capacitance of the series combination is given by ⇒ ⇒ Cnet = 3 x 10-12 F = 3 pico-Farad. (b) Let q be the charge on each capacitor. Sum of the potential difference across the plates of the three...
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Given, C1= C2= C3= 9 pF = 9 x 10-12F V = 120 volt a) Total capacitance of theseriescombination is given by ⇒ ⇒ Cnet= 3 x 10-12F = 3 pico-Farad. (b) Let q be the charge on each capacitor. Sum of the potential difference across the plates of the three capacitorsmustbe equal to 120 V ⇒V1+ V2+ V3= Vnet ⇒(∵Vnet= 120 V) ⇒ ⇒ ⇒q = 360 x 10-12C Therefore, potential difference across a capacitor read less
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Clear Your Concepts and Score Higher

(a)Total capacitance=C/3=9/3=3 pF using series capacitance formula (b)As capacitors are connected in series so each capacitor will hold equal amount potential. Hence p.d. across each capacitor is 120/3=40 V
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POST GRADUATE IN EEE, NOT HAVING EXPERIENCE IN TEACHING BUT HAVE EXPERIENCE IN PROBLEM SOLVING

a) Answer is 3pf because as the capacitors are connected in series. The formula is when there are 'n' identical capacitors each are having m' farads. Then the shortcut answer is m/n.for detailed explanation I will tell u when we come across together in sharing of knowledge(b). the Answer is 40v.as the...
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a) Answer is 3pf because as the capacitors are connected in series. The formula is when there are 'n' identical capacitors each are having m' farads. Then the shortcut answer is m/n.for detailed explanation I will tell u when we come across together in sharing of knowledge(b). the Answer is 40v.as the capacitors are identical and are in series the voltage across them will divide equally. (b). answer is 40v.as the capacitors are identical and are in series the voltage across them will divide equally. read less
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