3) connect two lamps to a power supply in series and current drawn from the power supply is is. connect the same two lamps in pa
1 answer:
The current drawn by the series circuit is(R₁ + R₂)/R₁R₂ times the current drawn by the parallel circuit.
Let the resistance of the two lamps are R₁ and R₂.
Then the equivalent resistance in series combination is: R = R₁ + R₂.
And, the equivalent resistance in parallel combination is:
r = R₁R₂/(R₁ + R₂).
So, if the supply voltage is V,
Then, current drown in series combination; = V/R = V/(R₁ + R₂)
And, current drown in parallel combination; = V/r = V(R₁ + R₂)/R₁R₂
So , = [ V/(R₁ + R₂)] /[V(R₁ + R₂)/R₁R₂]
= (R₁ + R₂)/R₁R₂
Hence, the ratio of current drawn in series and current drown in parallel is (R₁ + R₂)/R₁R₂. So, he current drawn by the series circuit is(R₁ + R₂)/R₁R₂ times the current drawn by the parallel circuit.
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Answer:
Explanation:
radius of the solenoid, r = 0.05 m
length of the solenoid, l = 0.39 m
Magnetic field of the solenoid, B = 2 x 10^-5 T
Number of turns, N = 200
The magnetic field of the solenoid is given by
where, i be the current and n be the number of turns per unit length
n = N / l = 200 / 0.39 = 512.8
i = 0.031 A
The spring constant is 4 N/m
Explanation:
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:
where
F is the magnitude of the spring applied
k is the spring constant
x is the elongation of the spring, relative to its equilibrium position
For the spring in this problem, we have:
F = 0.12 N (force applied)
x = 3 cm = 0.03 m (elongation of the spring)
Therefore, we can solve the formula for k to find the spring constant:
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