Answer:
In which direction does the current in circuit A flow?
counterclockwise
<h2>What is the power dissipated by the resistor of resistance R2 for circuit A, given that E=10 V, R1=300ohms, and R2=5000ohms? </h2><h2>Calculate the power to two significant figures.</h2><h2>0.064W</h2><h2 /><h2>For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?</h2><h2>R1/R2 = 1 </h2><h2 /><h2>Under which of the following conditions would power dissipated by the resistance R2 in circuit A be bigger than that of circuit B? </h2><h2>Some answer choices overlap; choose the most restrictive answer.</h2><h2>R2>R1</h2><h2> </h2>Explanation:
Answer:
The average induced emf in the loop is 0.20 V
Explanation:
Given:
Radius of loop m
Magnetic field T
Change in time sec
According to the faraday's law,
Induced emf is given by
Where magnetic flux
( here
)
Where
We neglect minus sign because it's shows lenz law
V
Therefore, the average induced emf in the loop is 0.20 V
Answer:
Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:
- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,
- Now for the limit y >>a:
- Insert limit i.e a/y = 0
Hence the Electric Field is off a point charge of magnitude 3q.