What is the resistance...

4. What is the resistance (R) of the resistor if the electric current (I) flowing through this resistor is equal to 2 A, and the electric potential (V_L) on the left terminal of this resistor is 4 V and the electric potential (V_r) on the right terminal of this resistor is 10 V. Which is direction of the electric current through this resistor, toward left or right?
V_r - V_L = ΔV_r,l = -I_r,l R
-I_r,l R = (V_r - V_L )

-I_r,l  = (V_r - V_L )/R

I_r,l  = (V_L – V_r )/R

 R = |(V_L – V_r )/ I_r,l |

R  = |(4V – 10V )/2A| = 3 Ω

The electric current flows from greater potential, 10V, right, to the lower potential, 4V, left: toward to left.

Exam 1. Problem 1

Exam 1. October 8, 2015, 8:00 AM
In solutions students should show:
a) given data by symbols,
b)   the unknown by symbol,
c) useful for this problem general formulae,
d) algebra job and final algebraic expression for the unknown,
e) substitution of symbols in the final algebraic expression by numbers with measurement units,
f) if it is required by a problem the final number with measurement units.

1. Electric field (E) accelerates a charged droplet from the zero speed to the final speed (V₁) of 8 m/s. What will be the final speed (V₂) if the electric field is decreased by a factor of 4, but all other parameters of this accelerator are not changed?

The electric field is decreased by a factor of 4 means: E₂= E/4, E₂/E=1/4

Potential energy of electric field, U=E∙d∙q
Kinetik Energy, K=½mv²

U=K

E∙d∙q = ½mv₁²,
where d is the traveling distance of the droplet in the electric field, q is the droplet’s electric charge.

E₂∙d∙q = ½mv₂²

(E₂∙d∙q)/ (E₁∙d∙q) =  (½mv₂²)/(½mv₁²)

E₂/ E =  v₂²/v₁²

v₂² = v₁² ∙E₂/ E

v₂ = v₁ ∙(E₂/ E)^1/2

v₂ = 8m/s ∙(1/4)^1/2 = 8m/s ∙(1/2) = 8m/s  /2 = 4m/s