{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Phy 121 - Sample Ex3a

# Phy 121 - Sample Ex3a - Phys 121 Exam 3 j =(ne vd j = i/A V...

This preview shows pages 1–2. Sign up to view the full content.

Phys 121 Exam 3 __________________________________________________________________________________________ j = (ne) v d j = i/A V = iR P = iV P = i 2 R = V 2 /R series: R eq = R 1 + R 2 +..R n parallel: R eq = (1/R 1 +1/R 2 +..1/R n ) -1 RC-circuit: charging: Q = Q (1-e -t/RC ), V C = V (1-e -t/RC ), i = i 0 (e -t/RC ), V R = i 0 R (e -t/RC ), discharging: Q = Q 0 e -t/RC , i = i 0 (e -t/RC ), V c = V 0 (e -t/RC ), V R = i 0 R (e -t/RC ), F m = q v x B F e = q E F m = i L x B v = E/B qΔ =K f -K I K=½mv 2 qvB = mv 2 /R R=mv/qB τ = μ x B μ = NiA B wire, out = μ 0 i/2 π R B wire, in = ( μ 0 i/2 π R 2 )r B arc = μ 0 i θ /4 π R B loop = μ 0 i/2R B solenoid = μ 0 in Flux: Φ = B A emf = -d Φ /dt emf = ( Φ 2 - Φ 1 )/ t emf mot = BLv emf max = NAB ϖ μ 0 = 1.26x10 -6 Tm/A e = 1.6x10 -19 C m e = 9.11x10 -31 kg 1 Two identical batteries of emf E=3V and internal resistance r=0.4- are connected in parallel then connected in series with R=2- resistor. The current in the 2- resistor is: A) 1.36 A B) 1.85 A C) 2.13 A D) 2.95 A E) 3.71 A 2. The power dissipated in the 2- resistor (in problem 2) is

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}