Answer:
current in series is 2.50 mA
current in parallel is 13.51 mA
Explanation:
given data
voltage = 5 V
resistors R1 = 1.5 kilo ohms
resistors R2 = 0.5 kilo ohms
to given data
current flow
solution
current flow in series is express as here
current = voltage / resistor .................1
put here all value in equation 1
current = 5 / (1.5 + 0.5)
current = 5 / 2.0
so current = 2.50 mA
and
current flow in parallel is express as
current = voltage / resistor ....................2
put here all value in equation 2
current = 5 / (1/ (1/1.5 + 1/0.5))
current = 5 / 0.37
so current = 13.31 mA
Answer:
The current drawn by Horace’s reading glasses is 0.8 A.
Explanation:
Given that,
Resistance of each bulb, R = 2 ohms
Voltage of the system, V = 3.2 volts
These two bulbs are connected in series. The equivalent resistance will be 2 ohms +2 ohms = 4 ohms
Let I is the current drawn by Horace’s reading glasses. Using Ohm's law to find it such that :
So, the current drawn by Horace’s reading glasses is 0.8 A.
Choice ' C ' is a true statement.
The other choices aren't.
The answer is D. I know because I already answered the question.
Answer:
a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d)
, c) m = 3
Explanation:
a) In the interference phenomenon the maxima are given by the expression
d sin θ = m λ
the maximum for m = 1 is at the angle
θ = sin⁻¹ λ / d
the second maximum m = 2
θ = sin⁻¹ ( λ / 2d)
the third maximum m = 3
θ = sin⁻¹ ( λ / 3d)
the fourth maximum m = 4
θ = sin⁻¹ ( λ / 4d)
b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.
I = I₀ cos² (Ф) (sin x / x)²
Ф = π d sin θ /λ
x = pi a sin θ /λ
where a is the width of the slits
with the values of part a are introduced in the expression and we can calculate intensity of each maximum
c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present
maximum interference d sin θ = m λ
first diffraction minimum a sin θ = λ
we divide the two expressions
d / a = m
In our case
3a / a = m
m = 3
order three is no longer visible
