Cr and lr circuits help exam in a bit and i'm stuck

[frixis]

cr and lr circuits help.. exam in a bit and I'm stuck!

okay so here's my question :
how is the peak voltage across the resistance (in a cr circuit) or the induction (in an rl circuit) = $$\pm$$V$$_{m}$$/(1-e$$^{-T/2\tau}$$)

V$$_{m}$$ is the max voltage of the pulse train.
the m denotes max

both circuits are in a steady state with a pulse response

if it were a positive step then v=V$$_{m}$$e$$^{-T/\tau}$$
and if it were a negative step it would be v=-V$$_{m}$$e$$^{-T/\tau}$$

so from this i can gather that the peak values will be a plus minus thing..
i can get to peak voltage = V$$_{m}$$-v
but apart from that I'm very very lost... I've been working on this for a long while now but...
help please

 

[KataKoniK]

Hi there,

I can definitely understand your confusion with this concept. It can be quite tricky to wrap your head around at first, but let me try to break it down for you.

In a CR circuit, the voltage across the resistance is given by the equation V = Vm(1-e^(-t/RC)). This means that as time goes on, the voltage across the resistance will approach the maximum voltage (Vm). However, it will never actually reach that maximum voltage, as indicated by the (1-e^(-t/RC)) term. This is because the capacitor in the circuit is charging up and limiting the current flow, causing the voltage to gradually increase.

Now, when we talk about peak voltage, we are referring to the maximum voltage that the circuit will reach. In the case of a CR circuit, this would be Vm, as that is the maximum voltage that the circuit is capable of producing. However, as we discussed earlier, the voltage across the resistance will never actually reach Vm. So, to calculate the peak voltage, we use the equation Vm - V, where V is the voltage across the resistance at any given time.

Similarly, in an RL circuit, the voltage across the inductor is given by V = Vm(1-e^(-t/LR)). Again, as time goes on, the voltage across the inductor will approach the maximum voltage (Vm), but will never actually reach it due to the inductor limiting the current flow.

So, to summarize, the peak voltage across the resistance (in a CR circuit) or the inductor (in an RL circuit) is equal to Vm - V, where V is the voltage across the resistance or inductor at any given time. This is because the maximum voltage (Vm) is never actually reached in these circuits, but rather approached asymptotically.

I hope this helps to clarify things for you. Best of luck on your exam!

 

[piercas]

I understand your frustration and confusion. The equations you are using are correct, but it is important to understand the underlying principles behind them. In a CR circuit, the peak voltage across the resistance is determined by the maximum voltage of the pulse train (V_m) and the time constant (τ). The time constant represents the time it takes for the voltage to reach 63.2% of its maximum value. This means that as time goes on, the voltage will approach but never reach the maximum value. This is why the equation for the peak voltage is V_m/(1-e^(-T/2τ)).

Similarly, in an LR circuit, the peak voltage across the inductance is determined by the maximum voltage of the pulse train and the time constant. However, in this case, the time constant represents the time it takes for the current to reach 63.2% of its maximum value. Again, the voltage will approach but never reach the maximum value. This is why the equation for the peak voltage is V_m/(1-e^(-T/2τ)).

In both cases, the plus/minus sign indicates that the peak voltage can be either positive or negative depending on the direction of the step. This is because in a CR or LR circuit, the voltage or current will change direction as the pulse train switches from positive to negative or vice versa.

I hope this explanation helps clarify the equations and principles behind them. Remember, it is important to understand the concepts rather than just memorizing equations. Keep up the good work and good luck on your exam!