Show step by step solution, no shortcuts.!!! SHOW ALL STEP BY STEP WORK AND EXPL

Question

Show step by step solution, no shortcuts.!!! SHOW ALL STEP BY STEP WORK AND EXPLANATION!!!!!!!!! STEP BY STEP

1. A 0.50-F and a 1.4-F capacitor (C1 and C2, respectively) are connected in series to a 17-V battery.

PART A: Calculate the potential difference across each capacitor.

PART B: Calculate the charge on each capasitor.

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

PART D: Calculate the charge on each capasitor assuming the two capacitors are in parallel.

Explanation / Answer

A&B. C1 = 0.50 uf

C2 = 1.4 uf

In series Ceq = C1*C2/(C1 + C2) = 0.5*1.4/(1.4+0.5) = 0.7/1.9 = 0.368 uf

emf = 17 V

Current is same in series connection.

Q1 = Q2 = Qeq = Ceq*V = 0.368*10^-6*17 = 6.26*10^-6 C

V1 = Q1/C1 = 6.26*10^-6/(0.5*10^-6) = 12.52 V

V2 = Q2/C2 = 6.26*10^-6/(1.4*10^-6) = 4.47 V

C&D.

Ceq = C1 + C2 = 1.4 + 0.5 = 1.9 uf

potential difference is equal in parallel connection

V1 = V2 = 17 V

Q1 = C1*V1 = 17*0.5*10^-6 = 8.5*10^-6 C

Q2 - C2*V2 = 17*1.4*10^-6 = 23.8*10^-6 C

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

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