Question

Did you dare touch the Van de Graaff? The maximum charge that our classroom Van
de Graaff generator could hold on its dome was about 5 µC. Assume that when you get
shocked by the generator, all of this charge passes through you in just one-hundredth
of a second (1.00 × 10−2
s). (a) Calculate the associated average current for getting
shocked by the Van de Graaff generator, and express your answer in units of mA
(milli-amperes). Now, let’s say you are feeling bold and opt to lick a 9 V battery. The
resistance for wet skin is about 2000 Ω. (b) Calculate the associated current through
your tongue when you place it across the terminals of a 9 V battery, and express your
answer in the same units of mA.

Answer 1

Answer on Question 65468, Physics, Electric Circuits

Question:

Did you dare touch the Van de Graaff? The maximum charge that our classroom Van de Graaff generator could hold on its dome was about $5\mu C$. Assume that when you get shocked by the generator, all of this charge passes through you in just one-hundredth of a second $(1.00 \cdot 10^{-2} s)$. (a) Calculate the associated average current for getting shocked by the Van de Graaff generator, and express your answer in units of $mA$ (milli-amperes). Now, let's say you are feeling bold and opt to lick a $9V$ battery. The resistance for wet skin is about $2000\Omega$. (b) Calculate the associated current through your tongue when you place it across the terminals of a $9V$ battery, and express your answer in the same units of $mA$.

Solution:

(a) By the definition of the current we get:

$$\bar{I} = \frac{\Delta Q}{\Delta t} = \frac{5 \cdot 10^{-6} \, C}{1.00 \cdot 10^{-2} \, s} = 0.5 \cdot 10^{-3} \, A = 0.5 \, mA.$$

(b) We can find the current through the tongue from the Ohm's law:

$$I = \frac{V}{R} = \frac{9 \, V}{2000 \, \Omega} = 4.5 \cdot 10^{-3} \, A = 4.5 \, mA.$$

Answer:

(a) $\bar{I} = 0.5 \, mA$.

(b) $I = 4.5 \, mA$.

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