Disha Karnataki
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hi everyone,
we were given a simple circuit as shown in the attachment, and we are asked to find out voltage across 5ohm resistance and then we can find out current.
my question is if by superposition theorem i find out effect of the voltage sources individually then:
1)potential difference(p.d.) of 2 ohm resistance due to 10v battery alone is 4.736885v so the potential at 5ohm resistance is 10-4.736885=5.26315v.
2)potential at 5 ohm sue to 5v acting alone is p.d. of 4ohm resistance is 3.6842v so the potential at 5ohm resistance is 5-3.6842=1.31578v.
i can find all this up-to here i have no doubt. then, it's mentioned that we have to find out total voltage across 5ohm so here i subtract the two potentials at that point i.e 5.26315v-1.31578v=3947337v. here my answer is wrong. because the correct answer is :5.26315v+1.31578v=6.57893v.
i am not understanding this how can two voltages here be additive? because u can see that 10v>5v hence,it's as if 10v is charging 5v & 5v is refusing to charge by imposing it's own effect,hence the voltages should be subtractive in nature.
we were given a simple circuit as shown in the attachment, and we are asked to find out voltage across 5ohm resistance and then we can find out current.
my question is if by superposition theorem i find out effect of the voltage sources individually then:
1)potential difference(p.d.) of 2 ohm resistance due to 10v battery alone is 4.736885v so the potential at 5ohm resistance is 10-4.736885=5.26315v.
2)potential at 5 ohm sue to 5v acting alone is p.d. of 4ohm resistance is 3.6842v so the potential at 5ohm resistance is 5-3.6842=1.31578v.
i can find all this up-to here i have no doubt. then, it's mentioned that we have to find out total voltage across 5ohm so here i subtract the two potentials at that point i.e 5.26315v-1.31578v=3947337v. here my answer is wrong. because the correct answer is :5.26315v+1.31578v=6.57893v.
i am not understanding this how can two voltages here be additive? because u can see that 10v>5v hence,it's as if 10v is charging 5v & 5v is refusing to charge by imposing it's own effect,hence the voltages should be subtractive in nature.