In this lab assignment, we will examine the natural response of a simple RC circuit. We will use both a manual switching operation and a square wave voltage source to create our circuit’s natural response. We will see that the method used to create the response affects the circuit being measured.
When we apply 5V power to the circuit, we got a graph, which is a linear part combined with a exponential part
In the lab, we find that the one period is 0.3678s because T=e^-1. And after one period, the value becomes 36.78% of its original value. and we find V0=3.432V, so after one period, the voltage is 1.260V. but When we find period, it is 49.5ms. its difference is -227%.
In part b, a 2.5 V square wave with low frequency is apply into circuit.
In part B, the max Voltage is 3.432V, and in same way, the experiment constant time is 15.25ms, find the different percent is -0.826%
Passive RL Circuit Natural Response
the circuit is set up.
Professor Mason did this lab for us.
Summary:Today we go over Capacitors and Inductors, and learn how to solve RC and RL circuit problems.
before we calculated the R1 and R2, and find C:
R1 R2
theoretical value: 1K ohm 2.2K ohm
experimental value: 0.98K ohm 2.14K ohm
the Value of C is 22uF.
Breakboard is set up for circuit.
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The graph about 2.5V square wave is applied.
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Analysis:
when we apply a square wave with a low frequency, its difference percent is low because during we apply a square wave, we do not to plug and unplug ourselves. It causes the high percent different 227%.
In this lab assignment, we will examine the natural response of a simple RL circuit. We will use both a manual switching operation and a square wave voltage source to create our circuit’s natural response. We will see that the method used to create the response affects the circuit being measured.
On pre-lab, we assume apply a square wave with amplitude 2.5V and offset 2.5V for the circuit. two constant time, one is 10ms and other one is 30ms.
the graph is made by professor Mason.
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