NEED HELP ON #7 AND #8 1) If you have five capacitors with capacitances 1.3 x 10
ID: 585480 • Letter: N
Question
NEED HELP ON #7 AND #8
1) If you have five capacitors with capacitances 1.3 x 10-6 F, 1.7 x 10-6 F, 4.2 x 10-6 F, and two 7.9 x 10-6 F in series. What is the equivalent capacitance of all five? C= .5409 x 10-6
2) Initially the capacitors are uncharged. Now a 13 V battery is attached to the system. How much charge is on the positive plate of the 4.2 x 10-6 F capacitor? Q= 7.02 x 10-6
3) What is the potential difference between the plates of the 4.2 x 10-6 F capacitor? V= 1.67 V
4) How much energy is stored in the entire capacitor system? PE= 45.63 x 10-6
5) If you have five capacitors with capacitances 1.3 x 10-6 F, 1.7 x 10-6 F , 4.2 x 10-6 F, and two 7.9 x 10-6 F in parallel. What is the equivalent capacitance of all five? C= 23 x 10-6
6) If one attaches a 13 V battery to the system, how much charge is on the positive plate of the 4.2 x 10-6 F capacitor? Q= 54.6 x 10-6
7) What is the potential difference between the plates of the 4.0 x 10-6 F capacitor? V= ____ V
8) How much energy is stored in the entire capacitor system? PE= _____ x 10-6 J
Explanation / Answer
According to the question we need to solve for 7) and 8) .
I am not able to see the capaciance c =4.0 x 10-6 F so I am solving for 4.2 x 10-6 F
7) What is the potential difference between the plates of the 4.0 x 10-6 F capacitor? V= ____ V
answer) v = q /c = 7.02 x 10-6 / 4.2 x 10-6 F = 1.6714285 Volts ===============ANSWER)
( NOTE : I HAVE SOLVED FOR CAPACITANCE = 4.2 x 10-6 F INSTEAD OF 4.0 x 10-6 F BECAUSE IT DOES NOT CONTAIN ANY CAPACITANCE WITH VALUE OF 4.0 x 10-6 F )
8) How much energy is stored in the entire capacitor system?
Answer) E = 1/2*Ceq*v^2 = 0.5 * 0.5409 x 10-6 * (13 ) ^2 = 45.70605 * 10^-6 Joule or ========answer)
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