1) A charge q = -1.9 ?C sitting in a uniform electric field described by the vec
ID: 2035946 • Letter: 1
Question
1) A charge q = -1.9 ?C sitting in a uniform electric field described by the vector E = 5.00 i + 4.80 j + 4.20 k (in V/m) is moved along the dotted path from point A (x=2.5, y=1.5, z=2.0) to point B (x=7.5, y=-4.5, z=9.0).
Calculate the change in potential for the charge.
For this circuit, the resistors used had values of; R1 = 3.50 k?, R2 = 820 ?, R3 = 4.60 k?, and ?1 = 1.2 V, ?2 = 1.2 V, ?3 = 12.0 V.
Find the value for the current i2 in the circuit.
3)
Two identical light bulbs each with resistance R are connected to identical batteries with emf V as shown below.
Which of the following statements is correct (Give ALL correct answers, eg. B, AC ...)
A) Each circuit will produce the same amount of light
B) Circuit B will produce the most light
C) Circuit B will produce the most light only if R is small
D) Circuit A will produce the most light only if R is small
E) Circuit B will produce the most light only if R is large
F) Circuit A will produce the most light
G) Circuit A will produce the most light only if R is large
Explanation / Answer
Change in potential = Displacement*Electric field
= { (7.5 - 2.5) i + (-4.5 - 1.5) j + (9 - 2) k }*(5i + 4.8 j + 4.2 k)
= (5 i - 6 j + 7 k)*(5i + 4.8 j + 4.2 k)
= 25 - 28.8 + 29.4
= 25.6 J
b)
Applying Kirchoff's Current Law on the left loop:
E1 = i1*R1 + i2*R2 + E2
So, 1.2 = i1*3500 + i2*820 + 1.2 -------- (1)
For the loop on the right:
E3 = -i3*R3 + i2*R2 + E2
So, 12 = -i3*4600 + i2*820 + 1.2
Now, i1 = i2 + i3
So, 12 = -(i1 - i2)*4600 + i2*820 + 1.2 ----------- (2)
Solving the two equations above simultaneously, we get :
i1 = -3.9*10^-4 A
i2 = 1.66*10^-3 A
c)
Power dissipated across the circuit A:
V^2/R + V^2/R = 2V^2/R
Now, for circuit B:
(V/2)^2/R + (V/2)^2*R = V^2/2R
So, options F and D are correct.
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