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1. A capacitor with low energy loss has a (A) high PF (B)low Q (C)high DF (D)low ESR?

D; low ESR (effective series resistance) It will have a low PF (power factor), high Q and low DF (dissipation factor)

1 Answers · Science & Mathematics · 20/08/2013

2. derive eq: p=pf (m/(m-ma))?

what is the variable ?

1 Answers · Science & Mathematics · 10/05/2009

3. 425 V is applied to a 7900 pF capacitor. How much energy is stored?

E = (1/2)CV^2 E = (1/2)(7900e-9)(425)^2 E = 0.71346875 J

3 Answers · Science & Mathematics · 15/02/2008

4. 600 V is applied to a 7000 pF capacitor. How much energy is stored?

Energy stored = (1/2) C V^2 C = 7000pF = 7000 * 10^-9 F V = 600V = (1/2)(7000*10^-9) (600*600) =1.26*10^9*10^-9 =1.26J EDIT : Yes I am sorry... 1 picofarad = 1*10^-12 farad....He is...

2 Answers · Science & Mathematics · 23/04/2008

5. jiofw[f[ew'mmmcwopfk[wf pcwq]pf?

i'm not doing your homework dude.

2 Answers · Science & Mathematics · 28/01/2009

6. 460 V is applied to a 6500 pF capacitor. How much energy is stored?

E = 0.5 CV^2 = 0.5*6500*10^-12*460*460 = 0.0006877 J = 6.877*10^-4 J

1 Answers · Science & Mathematics · 03/04/2013

7. 550 V is applied to a 6900 pF capacitor. How much energy is stored?

w=(c*v^2)/2 w= (6.9*10^-9*550^2)/2= 1.043625[mJ] or 1043.625[μJ]

1 Answers · Science & Mathematics · 06/02/2013

8. 450 V is applied to a 6700 pF capacitor. How much energy is stored?

W = (1/2)∙C∙V² = (1/2) ∙ 6700×10⁻¹²F ∙ (450V)² = 1.35675×10⁻³J = 1.35675mJ

1 Answers · Science & Mathematics · 22/03/2009

9. You need to construct a 100 pF capacitor?

ok, first things first, mm's to m's .2 mm = 2*10^-4 m the formula you need is : C = EA/d ( E is constant btw, it is the permittivity of free space and it equals 8.85*10^-12) so, now u just need to plug in you values: 100*10^-12...

2 Answers · Science & Mathematics · 14/02/2009

10. A 5.00 - pF parallel-plate air-filled capacitor with circular plates is to be used in a circuit...?

The capacitance of the two plates of area A and plate separation d is given by: C = εoA/d Then, A = Cd/εo = π r² and, r = √{Cd/πεo}, The potential difference across the plates...

3 Answers · Science & Mathematics · 30/09/2010