A lamp is rated 240 V, 60 W. Determine the resistance of the lamp when lit?
120\(\Omega\)
240\(\Omega\)
540\(\Omega\)
960\(\Omega\)
Correct answer is D
P = 60w, V = 240v, R = ?
P = \(\frac{V^2}{R}\)
: 60 = \(\frac{240^2}{R}\)
→ R * 60 = 57,600
R = \(\frac{57,600}{60}\)
R = 960\(\Omega\)
In a series R-L-C circuit at resonance, impedance is----
maximum
minimum
capacitive
inductive
Correct answer is A
Since the current flowing through a parallel resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its maximum value.
(Z=R )
The speed of fast-moving neutrons in a nuclear reactor can be reduced by using?
graphite rods
concrete shield
iron rods
boron rods
Correct answer is A
Moderation is the process of the reduction of the initial high speed (high kinetic energy) of the free neutron.
Since energy is conserved, this reduction of the neutron speed takes place by transfer of energy to a material called a moderator.
There are several different types of moderating materials, and each have places where they are used more effectively.
Typically-used moderator materials include heavy water, light water, and graphite.
0.200A
0.070A
0.050A
0.008A
Correct answer is D
Given Data: L = 10, V = 50, F = 100, I = ?
Inductive Reactance [X\(_L\)] = 2\(\pi\)FL → 2 * 3.14 * 100 * 10
X\(_L\) = 6,280
Current[I] = \(\frac{V}{X_L}\) → \(\frac{50}{6280}\)
I = 0.0079
≈ 0.008A
2.5\(\Omega\)
5.0\(\Omega\)
8.0\(\Omega\)
10.0\(\Omega\)
Correct answer is B
r = resistance of the galvanometer
l = current through galvanometer = 20mA or 0.02A
V1 = p.d across the galvanometer = I X r = 0.02 X r
: V1 = 0.02r
V2 = p.d across the multiplier = 8 - 0.02r
R = resistance of the multiplier = 395Ω
where R = \(\frac{V}{I}\)
--> 395 = \(\frac{8−0.02r}{0.02}\)
CROSS MULTIPLY
8−0.02r = 395 * 0.02
8−0.02r = 7.9
--> 0.02r = 8 - 7.9
∴ r = \(\frac{0.1}{0.02}\)
= 5Ω