A person with body resistance between his hands of 10KÙ accidently grasps the terminal of a 14-KV power supply.
(a) If the internal resistance of the power supply is 2000Ù, what is the current through the person’s body?
(b) What is the power dissipated in his body?
(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 1.00mA or less?
(a) I = V/R
where R = Rb + Ri
= (10 + 2)*10^3 = 12,000 ohms
Rb = body resistance
Ri = power supply internal resistance
(b) I^2*Rb
(c) Require that V/(Rb + Ri) < 10^-3 A
Solve for Ri
To find the current through the person's body, we can use Ohm's Law, which states that current (I) is equal to the voltage (V) divided by the resistance (R).
(a) We are given the voltage of the power supply (V = 14 kV) and the person's body resistance (R = 10 kΩ). However, we need to account for the internal resistance of the power supply (R_internal = 2 kΩ).
Since the person's body resistance is in parallel with the internal resistance of the power supply, we can use the formula for calculating the total resistance (R_total) in a parallel circuit:
1/R_total = 1/R + 1/R_internal
By substituting the given values, we can calculate the total resistance.
1/R_total = 1/10kΩ + 1/2kΩ
= (2 + 10) / (2 * 10)kΩ
= 12/20kΩ
= 0.6kΩ
Now, we can use Ohm's Law to find the current through the person's body:
I = V / R_total
= 14kV / 0.6kΩ
≈ 23.33mA
Therefore, the current through the person's body is approximately 23.33mA.
(b) To find the power dissipated in the person's body, we can use the formula for power (P) in a circuit:
P = I^2 * R
By substituting the values we obtained earlier:
P = (23.33mA)^2 * 10kΩ
= 23.33^2 * 10^-6 * 10^3
= 5.44W
Therefore, the power dissipated in the person's body is approximately 5.44W.
(c) To make the power supply safe by limiting the maximum current to 1.00mA or less, we need to find the maximum resistance the power supply should have.
We can rearrange Ohm's Law to find the resistance when the current is known:
R = V / I
By substituting the values we have:
R = 14kV / 1.00mA
= 14kV / 1.00 * 10^-3 A
= 14MΩ
Therefore, the internal resistance of the power supply should be at least 14MΩ for the maximum current in this situation to be 1.00mA or less.