During the lab, you will find the time-constant experimentally using both circui

Question

During the lab, you will find the time-constant experimentally using both circuit 2 and circuit 3. The moment at which you begin taking measurements of the decaying voltage (in circuit two) or decaying current (in circuit three) is not significant. To see this, suppose that the stopwatchii is started at time t1 0. Since Delta V(t) = [ Delta V(0)]e-t1 / tau then, Delta V(t) = [ Delta V(0)]e-(t-0) / tau = [ Delta V(0)]e-(t - t1 + t1) / tau = [[ Delta V(0)]e-t1 / tau] e-(t - t1) / tau. If we define ( Delta t)1 t - t1 then Delta V(t) = [[ Delta V(0)]e-( Delta t)1 / tau. But [ Delta V(0)]e-t1 / tau is juat a constant. The factor e-( Delta t)1 / tau is the only factor that depends on time. (Actually, [ Delta V(0)]e-t1 / tau is the voltage at a time t1 and is often called V(t1) or simply V1.) Will this decay still be an exponential function of time? Will the time-constant have changed? Explain. [2] The

Explanation / Answer

This will still be exponential, as it is exponential everywhere in the V-t graph.

The time constant is still the same, as it follows the same shape.

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