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Chapter 5

Q. 5.6

Power Shock

Suppose that a careless worker wires a transformer improperly and uses the wrong chemicals for its coolant. As a result, the worker creates a massive explosion that destroys the transformer.

The worker is fired and goes out panhandling, limping on a broken leg and wearing a sign, “will fix downed power lines for food.” The worker hitchhikes around the state looking for work until he finds the little white farmhouse in Figure 5.19.

“If you get those power lines out of my yard, I’ll give you a hamburger,” says the farmer. “And if you get them out of my fields, I’ll give you a whole cow. The one that got struck by lightning, that one over there, is lying under the wreckage of the barn.” The worker agrees, and goes to remove the lines from the farmer’s yard. After testing the lines to make sure the power was indeed off, he takes off his thick rubber gloves, because it was “too hot for gloves.”

Unfortunately, the careless worker forgot about the test signals that the power company occasionally sends through a disabled power line. Power companies sometimes send a test current through a disconnected line to check if it could be reconnected safely. The test signal checks if the problem has been resolved before power for that line is turned back on. The power company, ignorant of the fact that the storm has twisted and toppled their towers beyond repair, sends a 345-kV test signal through the lines.

The scenario is shown in Figure 5.20. The scenario of Figure 5.20 can be modeled by the circuit shown in Figure 5.21. This is indeed a simplified version of the power line circuits. The details of the equivalent circuits for a power line will be discussed in Chapter 9.

Calculate the time, t, that it will take for the worker to feel a lethal shock. Lethal shock occurs when the body current reaches i_{\text {body}} = 100 mA. See Chapter 15 for details on electric safety.

5.6
5.6-
5.6--

Step-by-Step

Verified Solution

Method 1: Applying KCL to node A in Figure 5.21, i_{v}+i_{c}+i_{\text {body }}=0. Now, because:

i_{v}=\frac{v_{C}(t)-345  kV}{1  k \Omega}, i_{c}=10  \mu F \frac{dv_{C}(t)}{dt}, \text { and } i_{\text {body }}=\frac{v_C(t)}{200  k \Omega}

the equation becomes:

\frac{v_{C}(t)-345  kV}{1  k \Omega}+10  \mu F \frac{dv_{C}(t)}{dt}+\frac{v_{C}(t)}{200  k \Omega}=0

After some mathematical manipulation:

\frac{dv_{C}(t)}{dt}+100.5  v_{C}(t)=34.5 \times 10^6

To find v_{C}(t), rearrange the equation:

\begin{aligned}\frac{dv_{C}(t)}{dt} &=-100.5\left(v_{C}(t)-\frac{34.5 \times 10^6}{100.5}\right) \\&=-100.5\left(v_{C}(t)-3.4328 \times 10^5\right)\end{aligned}

Solving for v_{C}(t), and using the fact that the initial voltage is zero:

v_{C}(t)=3.4328 \times 10^5\left(1-e^{-100.5 t}\right)

Note that v_{C}(t)=R_{\text {body }} i_{C}(t). Next, find the time when the current through the body is 100 mA. In this case, v_{C}(t)=200  k \Omega \times 100  mA=20  kV=20 \times 10^3. Replacing for v_{C}(t) :

20 \times 10^3=3.4328 \times 10^5\left(1-e^{-100.5 t}\right)

Now, the time can be found:

t=\frac{1}{-100.5} \ln \left(1-\frac{20 \times 10^3}{3.4328 \times 10^5}\right) \rightarrow t=597  \mu s

Does he have enough time to drop the wire? No.

Method 2: The voltage, v_{C}(t), can be found using the Thévenin equivalent:

\begin{aligned}v_{th} &=345 \times 10^3 \frac{200}{200+1}=3.4328 \times 10^5  V \\R_{th} &=\frac{200 \times 1}{200+1}=0.9950  k \Omega\end{aligned}

The time constant is:

\tau=R_{th} C=0.9950 \times 10^3 \times 10 \times 10^{-6}=\frac{1}{100.5}

The capacitor voltage can be written:

v_{C}(t)=v_{th}\left(1-e^{\frac{-t}{\tau}}\right)=3.4328 \times 10^5\left(1-e^{-100.5 t}\right)

and the solution can be completed as stated earlier.