Superposition Thereom

Thread Starter


Joined Nov 14, 2004
Can somebody please guide me of how I can obtain the voltage across the 25Ohm Resistor using the Superposition Thereom. I rewrote the circuit leaving one source every time but my answer does not seem to match with the lecturer's answer(-9.139V). I can't understand what I'm doing wrong, any help will be appreciated


Joined Nov 14, 2004
yes , the answer -9.13 is correct .

using superposition , the 3V source can be replaced by a source as the ckt is linear and the source has no internal resistance .

then we have a 20ohm // 25ohm combo .
// parallel combo :)
20//25 = 500/45 =11.11111111
so voltage across 20/25=V due to 14 V = 14*11.11111/(11.111111+5)
which is abt 9.65V(note the polarity of this voltage , if the nodes on the bottom are connected to the grnd assumption then this is -9.65 V)

now take the 3V source and replace 14 V by short

again a 5//25 ohm combo = 125 /30 = 4.16666

so voltage across that is 4.1666/(20+4.16666) * 3V which is abt .5 V

now this is +ive when the nodes at the bottom are at 0V (assumption)
so voltage across the 25 ohm is -9.65+.52 = -9.13

well , the values are approximated here ,but u will get the given ans

Thread Starter


Joined Nov 14, 2004
Thanks a lot, I solved it :D
my problem was that i didnt see 5 and 25 resistors in parallel when removing 14V source
once again many thanks