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miskamm [114]
4 years ago
12

the primary of a transformer connected to 170v has 10 turns. the secondary has 70 turns. find voltage. this is a step down trans

former
Physics
1 answer:
Nikolay [14]4 years ago
4 0
No it isn't. (Unless you connect it backwards.)

If the primary has 10 turns and the secondary has 70 turns,
then the voltage that appears across the secondary is
7 times the voltage that you feed to the primary.

If you're 'exciting' the primary with 170 volts, then you need
to be cautious around the secondary terminals, because there's 
1,190 volts there !

If you want to use your transformer in a step-down configuration,
you can certainly connect it up the other way around.

Feed the 170 volts to the winding with 70 turns.  You've reversed
the labels 'primary' and 'secondary', and that's OK.  The voltage
at the terminals of the 10-turn winding will be (170/7) = 24.3 volts.
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Answer : The power absorbed by the bulb is, 0.600 W

Explanation :

As we know that,

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Conversion used : (1 mA = 0.001 A)

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4 years ago
A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm
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Answer:

the thermistor temperature = 325.68 \ ^0 \ C

Explanation:

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A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K

Resistance of the thermistor R_1 = 20,000 ohms

Material constant \beta = 3650

Resistance of the thermistor R_2 = 500 ohms

Using the equation :

R_1 = R_2  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{R_1}{ R_2} =   \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

Taking log of both sides

In \ \frac{R_1}{ R_2} = In \  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})

\frac{1}{T_2} =   \frac{1}{T_1}  -          \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}

{T_2} =  \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}

Replacing our values into the above equation :

{T_2} =  \frac{3650*373}{3650 - In (\frac{20000}{500})373}

{T_2} =  \frac{1361450}{3650 - 3.6888*373}

{T_2} =  \frac{1361450}{3650 - 1375.92}

{T_2} =  \frac{1361450}{2274.08}

{T_2} = 598.68 \ K

{T_2} = 325.68 \ ^0 \ C

Thus, the thermistor temperature = 325.68 \ ^0 \ C

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