Power and Resistance in Electrical Circuits

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Suppose that a person has a resistance of 13.0 kiloohms as part of a circuit which passes through his hands. This person accidentally grasps the terminals of a power supply with a potential difference of 16.0 V.
PART A:If the internal resistance of the power supply is 2100 ohms , what is the current through the person's body?
I calculated the current to be 1.06×10−3 A by using the equation I = V/(R_p + R_i)
PART B: To find the power dissipated in his body I thought you use the equation P = I*V so P =(1.06*10^-3 A)*(16.0V) which gave me 1.70*10^-2 which is wrong. What I am doing wrong??
Thank you for your help
 
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You assumed that the 2 resistances are in series. Are they?
 
I noticed the same post in the Intro Physics section. You should avoid double posts.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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