Physics, 27.03.2020 04:11 spookymod8967

A transformer has 1000 turns in the primary coil and 100 turns in the secondary coil. If the primary coil is connected to a 120 v outlet and draws 0.050 a, what are the voltage and current of the secondary coil?1200v,0.0050 a1200v, 0.50a12v,00050a12v,0.50a

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Answer from: wavy419

12 V, 0.5 A

Explanation:

For the Voltage,

Using,

Vs/Vp = Ns/Np Equation 1

Where Vs = Voltage in the secondary coil, Vp = Voltage in the primary coil, Ns = number of turn in the secondary coil, Np = number of turns in the primary coil.

Make Vs the subject of the equation

Vs = Vp(Ns/Np) Equation 2

Given: Vp = 120 v, Np = 1000 turns, Ns = 100 turns

Substitute into equation 2

Vs = 120(100/1000)

Vs = 120×0.1

Vs = 12 v

For the current,

Using

Ns/Np = Ip/Is Equation 3

Where Ip = current in the primary coil, Is = current in the secondary coil

make Is the subject of the equation

Is = Ip(Np/Ns) Equation 4

Given: Np = 1000 turns, Ns = 100 turns, Ip = 0.05 A

Substitute into equation 4

Is = 0.05(1000/100)

Is = 0.05×10

Is = 0.5 A.

Hence the voltage and the current in the secondary coil is 12 V, 0.5 A

Answer from: anonymous1813

Option D) 12 V, 0.50 A

Explanation:

Number of turns in the primary coil, $N_{p} = 1000$

Number of turns in the secondary coil, $N_{s} = 100$

Voltage in primary coil, $V_{p} = 120 V$

Current drawn by primary coil, $I_{p} = 0.050 A$

Voltage in secondary coil, $V_{s} = ?$

Current in secondary coil, $I_{s} = ?$

Relationship between voltages applied in the secondary and primary coils

$\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }$

$\frac{V_{s} }{120 } = \frac{100 }{1000 }$

$V_{s} = 120*0.1\\V_{s} = 12.0 V$

Relationship between currents drawn by the secondary and primary coils

$\frac{I_{s} }{I_{p} } = \frac{N_{p} }{N_{s} }$

$\frac{I_{s} }{0.050 } = \frac{1000 }{100 }$

$I_{s} = 0.050*10\\I_{s} = 0.50 A$

Answer from: Quest

1.the balanced chemical equation is as below

cu + 2 ag(no)₃ → cu(no₃)₂ +2 ag

2.the much silver produced is 50.9 g

step 1: find the moles of cu

moles = mass÷ molar mass

from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236 moles

step 2: use the mole ratio to determine the moles of ag

from equation above cu: ag is 1: 2

therefore the moles of ag =0.236 moles x 2/1 =0.472 moles

step 3: find the mass of ag

mass = moles x molar mass

from periodic table the molar mass of ag =107.87 g/mol

mass = 0.472 moles x 107.87 g/mol = 50.9 g

3. the much silver collected worth is 1.797 oz

convert 50.9 g to oz

that is 1 gram = 0.0353 oz

50.9 grams =? oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1 gram} =1.797 oz

Answer from: Quest

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.

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A transformer has 1000 turns in the primary coil and 100 turns in the secondary coil. If the primary...

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