Adding I1,I2,I3,I4 to Get I5 - Tips & Tricks

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I5 is calculated as I3 - I4 + I2 + I1, confirming the conservation of electric charge. The discussion emphasizes that current should be viewed as a flow, where the total current at a point is the sum of incoming currents minus outgoing currents. It is clarified that all currents entering a junction are considered positive, while those leaving are negative. The final value for I5 is determined to be -12A, adhering to the established rules of current flow. Understanding these principles is essential for accurate calculations in electrical circuits.
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What is I5 I think its I3-I4+I2+I1 but not surehow to add em up any tips?

thanks
 

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Yes,it is correct.Electric charge is conserved.With the regular choise of signs,the I_{5} should be -12A.

Daniel.
 
Matty9879 said:
What is I5 I think its I3-I4+I2+I1 but not surehow to add em up any tips?
Think of current as water flow. The current at a point is the sum of all currents entering at that point minus all currents leaving that point. Your answer is correct.

AM
 
Thanks for the reassurance but by the way it wasnt neative
 
It had to be.That's the rule.All currents entering a vertex are positive,while all which are leaving are negative.

Daniel.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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