Internal resistance seen by the voltage source, Vx?

AI Thread Summary
The discussion focuses on calculating the resistance seen by a voltage source in an op-amp configuration, with the expected answer being 3 kΩ. The user seeks clarification on how to arrive at this value. Participants suggest marking the current direction and magnitude in the circuit to better understand the relationships between the resistors. The conversation emphasizes the importance of analyzing the circuit systematically to determine the equivalent resistance. Overall, the thread aims to clarify the calculation process for the resistance in the given op-amp setup.
Kratos321
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Hi everyone,

I have attached an image of the op-amp configuration (all op-amps are ideal). Basically, the question asks to calculate the resistance seen through the voltage source. The answer is 3 kΩ but I just don't understand how this is found. If someone could assist me, I would really appreciate it.

(By the way, I mean just resistance in the title, not internal resistance)

Cheers
 

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You have labelled the current flowing in that top 1k resistor as Ix. Using the notation you have established, now mark in the magnitude and direction of the current in each of the other 4 input resistors.
 
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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