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1. Homework Statement
hello Everyone, I've attached the Circuit to this post. We are suppose to solve it using superposition. we are suppose to find the current through the 9 ohm resistor.
2. Homework Equations
Some relevant equations I thought were the basic ones to solve circuits, for example: Ohm's law, which stats V=IR
3. The Attempt at a Solution
The thing that is throwing me off with this particular question is that there is a circle thing that has 5A, with an arrow going up. I have NO idea what that arow means, I even googled. The way I thought we would solve this would be to find the Rseries, so would I just add up all the resistances to find that? i.e would I add 4+6=10, then i'd do 1/10 +1/9 (because they are parallel) and then inverse it to equal 4.74, which we would then add to 3, and then do 1/7.74+1/20 (because they are parallel) to obtain 5.58 as a final resistance? I know that superposition means we have to answer the question from both sides. But i don't understand how we solve it from the side of the 5A.
for the 10V part, do we just do V=IR and then do 10V= 5.58I so obtain I=1.79?
thanks for the help!!!
hello Everyone, I've attached the Circuit to this post. We are suppose to solve it using superposition. we are suppose to find the current through the 9 ohm resistor.
2. Homework Equations
Some relevant equations I thought were the basic ones to solve circuits, for example: Ohm's law, which stats V=IR
3. The Attempt at a Solution
The thing that is throwing me off with this particular question is that there is a circle thing that has 5A, with an arrow going up. I have NO idea what that arow means, I even googled. The way I thought we would solve this would be to find the Rseries, so would I just add up all the resistances to find that? i.e would I add 4+6=10, then i'd do 1/10 +1/9 (because they are parallel) and then inverse it to equal 4.74, which we would then add to 3, and then do 1/7.74+1/20 (because they are parallel) to obtain 5.58 as a final resistance? I know that superposition means we have to answer the question from both sides. But i don't understand how we solve it from the side of the 5A.
for the 10V part, do we just do V=IR and then do 10V= 5.58I so obtain I=1.79?
thanks for the help!!!
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