Solving a Capacitance Problem: Finding the Voltage Drop Across a Resistor

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The discussion centers on a problem involving a 6V DC voltage source connected to a 5kohm resistor and a 10 nF capacitor in series. The main question is about determining the voltage drop across the resistor. Clarification is needed regarding whether the voltage drop is being asked for immediately after the connection or after some time has passed, as the answers differ significantly. The initial state requires understanding the behavior of the circuit at the moment of connection, while the final state involves the capacitor charging. The conversation emphasizes the importance of distinguishing between these two scenarios for accurate problem-solving.
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Homework Statement


A 6V DC voltage source is connected across both a 5kohm resistor and a 10 nF capacitor in series. The voltage drop across the resistor is what?
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Homework Equations



The Attempt at a Solution


should the charge=0 sicne by connecting wires across the capacitor you are discharging it
 
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Your "attempt" is not a valid line of reasoning. Try again, but first:

Does the question indicate whether the initial state (immediately after the connection) or the final state (after "some time" has passed) is wanted? The answers are completely different.

Answer the answers to either state requires no calculation.
 
You posted the same question 3 times. Please stop doing this
 
it is immediately after the connection
 
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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