A 0.50- F and a 1.4- F capacitor ( C 1 and C 2, respectively) are connected in s

Question

A 0.50-F and a 1.4-F capacitor (C1 and C2, respectively) are connected in series to a 7.0-Vbattery.

a. Calculate the potential difference across each capacitor.

Express your answers using two significant figures separated by a comma.

b. Calculate the charge on each capasitor.

Express your answers using two significant figures separated by a comma.

c. Calculate the potential difference across each capacitor assuming the two capacitors are in parallel.

Express your answers using two significant figures separated by a comma.

d. Calculate the charge on each capasitor assuming the two capacitors are in parallel.

Express your answers using two significant figures separated by a comma.

Explanation / Answer

a) When the capacitors are in series, the equivalent capacitance =C= C1C2/(C1+C2)=(0.5*1.4)/(0.5+1.4)=0.368 micro Farad

Charges on each capacitor = Q=0.368*7 = 2.576*10-6 Coulomb

Potential difference across C1 = (2.576*10-6)/(0.5*10-6)=5.152 V

Potentail difference across C2 = (2.576*10-6)/(1.4*10-6)=1.84V

b) as calcultaed above, charge on each capacitor = 2.576*10-6 Coulomb

c)When the capacitors are in parallel, potential drop across each capacitor=supply voltage=7 V

d)Charge on C1 = (0.5*10-6)*7 = 3.5*10-6 Coulomb

  Charge on C2 = (1.4*10-6)*7 = 9.8*10-6 Coulomb

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.