The current I through the 40 k resistor in the figure below

The current (I) through the 4.0 k resistor in the figure below is 3.50 mA. What is the terminal voltage V_ba of the \"unknown\" battery? (There are two answers. Why?)

Solution

Assuming I is to the right of 4 Kohms resistor is

Voltage across 4 kohms is

V₄=IR=4000*(3.5*10^-3)=14 Volts

Current through 8kohms resistor is

I₈=V₄/R =14/8000 =1.75 mA

So total currentis

I =3.5+1.75 =5.25 mA

Applying KVL around outer loop,Assuming we go clockwise around the loop ,

V_ba-IR₅-I₄R₄-E-Ir =0

V_ba =(5.25*10^-3)*5000 + (3.5*10^-3)*4000+12+(5.25*10^-3)*1

V_ba =52.255 Volts

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Assuming I is to the left of 4 Kohms resistor is

If we go counterclockwise around the outer loop ,

-Ir +E-I₄R₄-IR₅+V_ab=0

=>V_ab =(5.25*10^-3)*1 -12+(3.5*10^-3)*4000+(5.25*10^-3)*5000

V_ab =28.255 Volts

=>V_ba=-V_ab =-28.255 Volts