In a circuit, three pieces of wire labeled A, B and C are joined at a common point, D. If wire B carries 1.5 mA in a direction away from D and C carries current 1.3 mA toward D, what does wire A carry?

a. 0.2 mA toward D
b. 0.2 mA away from D
c. 2.8 mA toward D
d. 2.8 mA away from D
e. unable to determine

correct option is a. 0.2 mA toward D

Explanation:

given data

B carries = 1.5 mA

C carries current  = 1.3 mA

solution

we take positive direction of current going away from the point D

and negative direction of current coming towards point D

so we use here kirchoff's current law   that is

iA + iB + iC = 0    ......................1

iA + 1.5 + (-1.3) = 0

iA = - 0.2 mA  

so that current in wire A is 0.2 mA towards point D

correct option is a. 0.2 mA toward D

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards