Capacitor and Resistance in a Series Circuit: Effects on Ammeter Reading?

AI Thread Summary
In a series circuit with a battery, resistor, capacitor, and ammeter, closing the switch initiates the charging of the capacitor. As the capacitor charges, its resistance effectively increases, leading to a gradual decrease in the ammeter reading. The current diminishes until the voltage across the capacitor matches the source voltage. This behavior illustrates the relationship between capacitance, resistance, and current in a circuit. Understanding this dynamic is crucial for analyzing series circuits effectively.
Peter G.
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There is a series circuit with a battery, resistor, capacitor and ammeter and a switch.
The question asks me what will happen to the ammeter reading when the switch is closed and why:

Well, I know that a capacitor charges up. What I am thinking is that as it charges up, it's resistance increases and therefore, the ammeter reading gradually decreases. Is that correct?
 
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As the capacitor charges the potential in the capacitor increases until it matches that of the source. You are quite right in saying that the current gradually decreaes.
 
Ok, cool! Thanks a lot Kurdt. :smile:
 
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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