A person with body resistance between his hands of 10k? accidentally grasps the

Question

A person with body resistance between his hands of 10k? accidentally grasps the terminals of a 14-kV power supply.

Part A

If the internal resistance of the power supply is 1800? , what is the current through the person's body?

Express your answer using two significant figures.

Part B

What is the power dissipated in his body?

Express your answer using two significant figures.

Part C

If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be Imax=1.00mA or less?

Express your answer using two significant figures.

Explanation / Answer

V = 10 KV = 10000V

Net resistance = 10000 + 1800 = 11800 ohm

a)So current ,I = 10000/11800 = 0.85 A (ans)

b) Power dissipated ,P =I^2R = 0.85*0.85*10000 = 7225 W (ans)

c)Let internal resistance =r

I = V/(R+r) = 10000/(10000 + r) <=0.001

10 + 0.001r > = 10000

0.001r> = 9990

r>=9990000 ohm

Minimum value of internal resistance = 9990000 ohm =9990 Kohm (ans)

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