Resistance and current question

[Gardalay]

Homework Statement

Suppose that a person's body resistance is 900ohms.
a) what current passes through the body when the person accidentally is connected to 110V?
b) if there is an alternative path to ground whose resistance is 40 ohms, what current passes through the person?
c) if the voltage source can produce at most 1.5A, how much current passes through the person in case b)?

Homework Equations

V=IR

The Attempt at a Solution

A) V=IR, I=0.122
B)1/900+1/40=0.026^-1=38.29 ohms, 110/38.29=2.87amps, 0.122a passes through the person, 2.75amps passes through the ground

C) Would the answer just be 1.372amps since 2.75-1.5=1.25 and then 1.25+0.122=1.372?

 

[pgardn]

Homework Statement

Suppose that a person's body resistance is 900ohms.
a) what current passes through the body when the person accidentally is connected to 110V?
b) if there is an alternative path to ground whose resistance is 40 ohms, what current passes through the person?
c) if the voltage source can produce at most 1.5A, how much current passes through the person in case b)?

Homework Equations

V=IR

The Attempt at a Solution

A) V=IR, I=0.122
B)1/900+1/40=0.026^-1=38.29 ohms, 110/38.29=2.87amps, 0.122a passes through the person, 2.75amps passes through the ground

C) Would the answer just be 1.372amps since 2.75-1.5=1.25 and then 1.25+0.122=1.372?

I would solve part C differently. 1.25 amps is only a little bit more than 10X as much current as 0.122 amps, yet one resistor has over 20X the resistance compared to the other. So the difference in current between the body and the other path must be greater than what you have and still equal to a total of 1.5 amps when added toegther.

 

[Gardalay]

I found out the answer was to be 1.5*40/940.
I have no idea how the person got 40/940 can someone please explain

 

[berkeman]

I found out the answer was to be 1.5*40/940.
I have no idea how the person got 40/940 can someone please explain

It's just from Ohm's Law. Have you learned about voltage dividers yet? How about current dividers? The problem presents you with a current divider. If you have the same voltage across two parallel resistors, how does the current divide between the two paths? How would you calculate it?

 

[Gardalay]

Wouldn't I just add 1/40+1/900=38.29ohms?

 

[berkeman]

Wouldn't I just add 1/40+1/900=38.29ohms?

Not if you want to figure out how the current divides between the two branches, which I think is part of this question.

If you have 1 Ohm in parallel with 2 Ohms, how does the current divide between the two branches? In what ratio?