P 1.7-3 The element currents and voltages shown in Figure P 1.7-3 are correct with one exception: the reference direction of exactly one of the element currents is reversed. Determine which reference direction has been reversed.

 

c01f029

Figure P 1.7-3

First identify the six elements.

 

We are going to calculate power and see if the total adds to zero.  This time not all elements are using the Passive Convention (current goes from + terminal of an element to the – terminal).  So those that do follow this are absorbing power and we will calculate them as –iV.  For those that do not follow this, we will assume power is being delivered and we will calculate power as iV.

 

Element

-iV

P

A

-(-3A)(3V)

+9 W

 

 

 

B

(4A)(1V)

+4 W

 

 

 

C

-(-2 A)(2 V)

+4 W

 

 

 

D

(7 A)(5 V)

+35 W

 

 

 

E

-(2A)(-6 V)

+12 W

 

 

 

F

-(-5 A)(-8 V)

-40 W

 

 

 

Total

 

64 W – 40 W = +24 W

 

 

 

 

So we know at least one element is improperly labeled.  So as we reverse a current, the effect will change the power by a factor of 2.  So dividing 24 W by 2 gives us 12 W needs to be reversed, so the simpliest switch would be to reverse the current on Part E.  If we replace 2 A with -2 A, we would get -12 W which will lead to 52 W – 52 W which equals zero.

 

Change the current between nodes b and d (what I call part E)

from 2 A to -2 A!

 

 

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This page last updated on December 26, 2018