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1 year ago

Please help

Physics

2 answers:

vladimir2022 [97]1 year ago

5 0

Answer:

true true false true false

Greeley [361]1 year ago

4 0

Answer:

1)true

2)true

3)false

4)false

5)true

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2 years ago

An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current

balu736 [363]

Answer:

(a) The resistance R of the inductor is 2480.62 Ω

(b) The inductance L of the inductor is 1.67 H

Explanation:

Given;

emf of the battery, V = 16.0 V

current at 0.940 ms = 4.86 mA

after a long time, the current becomes 6.45 mA = maximum current

Part (a) The resistance R of the inductor

$R = \frac{V}{I_{max}} = \frac{16}{6.45*10^{-3}} = 2480.62 \ ohms$

Part (b) the inductance L of the inductor

$\frac{Rt}{L} = -ln(1-\frac{I}I_{max}})\\\\L = \frac{Rt}{-ln(1-\frac{I}I_{max})}}$

where;

L is the inductance

R is the resistance of the inductor

t is time

$L = \frac{Rt}{-ln(1-\frac{I}I_{max})}} = \frac{2480.62*0.94*10^{-3}}{-ln(1-\frac{4.86}{6.45})} \\\\L =\frac{2.3318}{1.4004} = 1.67 \ H$

Therefore, the inductance is 1.67 H

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3 years ago