Why is there no current in the right loop of this circuit?

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The discussion centers on why there is no current in the right loop of a specific circuit involving an inductor. It explains that immediately after the switch closes, the current is zero because the inductor's current is defined as an integral that starts from zero when the switch was open. The formula indicates that at the moment the switch closes, the initial current (i(t_0)) is still zero. This results in no current flowing through the right loop at that instant. Understanding this concept is crucial for analyzing circuits with inductors.
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Because for any inductor the current is an integral:
i = \int_{t_0}^{t}\frac{v(t)}{L}\ dt + i(t_0)

Immediately after the switch S closes mean that t = t_0, so the current is zero.

i(t_0) is zero since before that moment the switch was open.
 
I don't understand still...
 
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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