Find the voltage in the middle of this circuit?

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Steps are appreciated.

1 Answer
Ananda Dasgupta
Apr 5, 2018

#-33.75" V"#

Explanation:

The current of #18# mA splits into the two parallel branches of resistance (10 k+15 k) and (3 k+12 k) respectively, in inverse ratio of their resistances. Hence the currents are

#18" mA" times 15/(15+25)qquad "and"qquad 18" mA" times 25/(15+25)#
respectively.

Thus, the voltage drops in the 15 k and 12 k resistors are

#18" mA" times 15/(15+25) times 15" k" = 101.25" V"#

and

#18" mA" times 25/(15+25) times 12" k"= 135" V"#

Since there are no voltage drops across the 2 k and 4 k resistors, we have

#v_o = 101.25" V"-135" V" = -33.75" V"#

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