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2 years ago

Why there is no inductive reactance in dc circuits

1 answer:

4 0

when voltage source is connected across an ideal inductor, the current through it is limited by self induced EMF( N.d(phi)/dt).. This EMF opposes its cause by Lenz's law.

When steady dc supply is connected, there is no change of flux and hence no induced EMF is developed and inductive reactance offered to steady DC supply is zero.

At the same time this induced EMF is proportional to rate of change of magnetic flux/voltage, hence the inductive reactance offered is also proportional to frequency of voltage source.

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A square loop of side length a =4.5 cm is placed a distance b = 1.1 cm from a long wire carrying a current that varies with time

sergey [27]

Answer:

a)Current will flow perpendicularly.

b)Magnitude of flux will be 2.987 N m2 C−1

3 0

2 years ago

Read 2 more answers

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

mote1985 [20]

Answer:

Frequency = 658 Hz

Explanation:

The third harmonics of A is,

$f_{A}=(3)(440Hz)=1320Hz$

The second harmonics of E is,

$f_{E}=(2)(659Hz)=1318Hz$

The difference in the two frequencies is,

delta_f = 1320 Hz - 1318 Hz = 2 Hz

The beat frequency between the third harmonic of A and the second harmonic of E is,

delta_f = $3f_{A}-2f_{E}$

$f_{E}=\frac{3f_{A}-delta_f}{2}$

We have calculate the frequency of the E string when she hears four beats per second, then

delat_f = 4 Hz

$f_{E}=\frac{3(440Hz)-4Hz}{2}$

$f_{E}=658Hz$

Hope this helps!

5 0

3 years ago

A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of

Blizzard [7]

Answer:

the strength of the magnetic field is 3 x 10⁻⁵ T

Explanation:

Given;

velocity of the cosmic ray, v = 5 x 10⁷ m/s

force experienced by the ray, f = 1.7 x 10⁻¹⁶ N

angle between the ray's velocity and the magnetic field, θ = 45⁰

The strength of the magnetic field is calculated as;

$F = qvB \ sin(\theta)\\\\B = \frac{F}{qv\times sin(\theta)} \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 \times 10^{-19} \ C\\\\B = \frac{1.7\times 10^{-16}}{(1.602 \times 10^{-19})\times (5\times 10^7) \times sin \ (45)} \\\\B = 3 \times 10^{-5} \ T$

Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T

7 0

2 years ago

) A 73-mH solenoid inductor is wound on a form that is 0.80 m long and 0.10 m in diameter. A coil having a resistance of is tigh

Aleonysh [2.5K]

Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil

Answer:

6.169 μA

Explanation:

Formula for induced EMF is given by the equation;

EMF = M(di/dt). We are given;

di/dt = 2.5 A/s

M = 19μH = 19 × 10^(-6) H

Thus;

EMF = 19 × 10^(-6) × 2.5.

EMF = 47.5 × 10^(-6) V

Formula for current is;

i = EMF/R. R is resistance given as 7.7 ohms.

Thus; i = 47.5 × 10^(-6)/7.7

i = 6.169 μA

5 0

2 years ago

I can not find the answer to the one after potassium ?

Zepler [3.9K]

Answer:

it’s halogen

Explanation:

3 0

2 years ago