Find the equivalent resistance from the indicated terminal pair of the networks in the attached doc

a) R = 2.5 Ω, b) R = 1 Ω, c)     R = 2R / 3 Ω

Explanation:

The resistance configuration can be in series or in parallel, for each one the equivalent resistance can be calculated

series, the equivalent resistance is the sum of the resistances

parallel, the inverse of the equivalent resistance is the inverse of the sum of the resistances

let's apply these principles to each case

case a)

    equivalent series resistance

         R₁ = 1 +4 = 5 ohm

         R₂ = 2 +3 = 5 ohn

these two are in parallel

        1 / R = 1/5 +1/5

        1 / R = 2/5

         R = 2.5 Ω

case B

we solve the parallel

       1 / R₁ = ½ + ½ = 1

        R₁ = 1 Ω

we solve the resistors in series

       R₂ = 1 + 1

       R₂ = 2 Ω

finally we solve the last parallel

       1 / R = ½ +1/2 = 1

        R = 1 Ω

case C

we solve house resistance pair in series

     R₁ = R + 2R = 3R

we go to the next mesh

     R₂ = R + 2R = 3R

     R₃ = R + 2R = 3R

last mesh

     R₄ = R + R = 2R

now we solve the parallel of this equivalent resistance

     1 / R = 1 / R₁ + 1 / R₂ + 1 / R₃ + 1 / R₄

     1 / R = 1 / 3R + 1 / 3R + 1 / 3R + 1 / 2R

      1 / R = 3 / 3R + 1 / 2R = 1 / R + 1 / 2R

     1 / R = 3 / 2R

      R = 2R / 3 Ω

momentum of soft ball three times less than baskey ball

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s