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Helen [10]
3 years ago
5

after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF

capacitor from a camera flash unit retains a voltage of 150 V when an unwary student removes it from the camera. If the student accidentally touches the two terminals with his hands, and if the resistance of his body between his hands is 1.8 kΩ, for how long will the current across his chest exceed the danger level of 50 mA?
Physics
1 answer:
Snezhnost [94]3 years ago
4 0

Answer:

The time is 110.16\times10^{-3}\ sec

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

i_{0}=\dfrac{V_{0}}{R}

Put the value into the formula

i_{0}=\dfrac{150}{1.8\times10^{3}}

i_{0}=83.3\times10^{-3}\ A

We need to calculate the time

Using formula of current

i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}

Put the value into the formula

50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}

\dfrac{50}{83.3}=e^{\frac{-t}{RC}}

\dfrac{-t}{RC}=ln(0.600)

t=0.51\times1.8\times10^{3}\times120\times10^{-6}

t=110.16\times10^{-3}\ sec

Hence, The time is 110.16\times10^{-3}\ sec

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Explanation:

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An electric dipole consists of charges +2e and -2e separated by 0.82 nm

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Distance between charges, d=0.82\ nm=0.82\times 10^{-9}\ m

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When dipole moment is parallel to the electric field, \theta=0

\tau=p\times E\ sin(0)

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Hence, this is the required solution.

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