What is the RMS Voltage Across a Body Near a Transformer?

AI Thread Summary
The discussion focuses on calculating the RMS voltage across a person's body near a transformer, given a secondary voltage of 6000 V and stray capacitance of 10.0 pF. Participants suggest redrawing the circuit to simplify the secondary transformer as an AC source and using formulas to find the current and impedance. The impedance is calculated using the formula Z = 1/(jωC + 1/R), and Ohm's law is applied to find the current. Ultimately, the maximum voltage across the body is derived, and the RMS voltage is calculated using the relationship Vrms = Vmax/sqrt(2). The final calculated RMS voltage across the body is approximately 23 volts, indicating a low but potentially hazardous current flow.
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A person is working near the secondary of a transformer, as shown in the figure below. The primary voltage is 120 V at 60.0 Hz. The capacitance C, which is the stray capacitance between the hand and the secondary winding, is 10.0 pF. Assuming that the person has a body resistance to ground R, determine the rms voltage across the body. (Suggestion: Redraw the circuit with the secondary of the transformer as a simple AC source.)
FIGURE: http://east.ilrn.com/graphing/bca/user/appletImage?dbid=445132339

I need some help with this stuff. This is what I know already. I know that the secondary voltage is 6000 V according to the figure. Also, I know that the rms voltage = max voltage/sqrt(2). I'm not sure how to get the max voltage. I tried using the secondary voltage as the max voltage, but the answer wasn't right. So, I think the secondary voltage is the AC source. How do I get the max voltage using the secondary voltage? Maybe I'm doing this all wrong? I don't know...any help would be great! Thx in advance! :)
 
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How about trying this? Let the voltage (secondary) as

v=v_0 e^{jwt}

Find the current through the circuit i, using ohms law.

v_0 e^{jwt} = i (R_b -\frac{j}{wc})

where -j/wc is the impedance of the capacitor.

Find the charge of the capacitor q by integrating current i since
i = dq/dt

use q= Cv to find the voltage across the capacitor v_c

therefore, voltage across the body v_b = v - v_c

absolute value of the vb is the max voltage created across the person's body. I am getting this to be about 23 volts using the human body resistance of 1 Mega ohm, causing a current of about 0.023 mA to flow through him. Hopefully it would not kill him. :smile:
 


Hi there,

First of all, let's redraw the circuit as suggested. We can simplify the secondary of the transformer as a simple AC source with a voltage of 6000 V. The stray capacitance C can be represented as a capacitor in parallel with the body resistance R, which is connected to ground.

Now, we can use the formula V = IZ to find the voltage across the body, where Z is the impedance of the circuit and I is the current flowing through it. We can find the impedance by using the formula Z = 1/(jωC + 1/R), where j is the imaginary unit and ω is the angular frequency (2πf).

So, plugging in the values we have, we get Z = 1/(j2π(60)(10x10^-12) + 1/R) = 1/(j0.012 + 1/R). Now, we need to find the current I. We can use Ohm's law, V = IR, to find the current. Since we know the voltage (6000 V) and the resistance (R), we can solve for I.

Finally, we can use the formula V = IZ to find the voltage across the body, which will give us the maximum voltage. We can then use the formula Vrms = Vmax/sqrt(2) to find the rms voltage across the body.

I hope this helps you understand how to approach this problem. Let me know if you have any further questions. Good luck!
 
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