Question

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[Resistance-Capacitance] = [Resistance] middot [Capacitance] = volta/amp middot coulomb/volt = coulomb/volt = change/change/time = time. Thus, the dimensions of RC are the dimensions of time. Since RC is also a constant, tau = RC is called the 'time constant' of the circuit.) This constant is an important one: it controls the rate of discharge of the capacitor (through the resistor). Because delta V(0) = RI, the voltage across the capacitor is delta v(t) = RI_0e^t/tau delta V(0)e^-t/tau (where delta V(0) RI_0). Delta V(t) = delta V(0)e^-t/tau doesn't look as neat as I(t) = I_0e^-t/tau but it is the relation that you'll verify experimentally. How can you change the time constant if the capacitance of the circuit is fixed? [1]. If we succeed in measuring the time constant during the experiment, we'll have verified that delta V(t) = delta V(0)e^-t/tau. The relation delta V(t) = delta V(0)e^-t/tau says that the voltage drops asymptotically to zero as t rightarrow infinity. We might ask the question: "how long do we have to wait until the voltage across the capacitor is zero?" but the answer will be "After an infinite time". Initially. how large is delta V(t)/delta V(0)? After an infinitely long time, what is the value of delta V(t)/delta V(0)? [2] Saying that "the time t for delta V(t)/delta V(0) to be zero will be t = infinity" is always true but doesn't allow us to see the fact that some voltage decays are very fast and others are slower. We can characterize the rate of voltage decay by saying how long we have to wait for delta V(t) to become a certain fraction of its initial value. To pursue this, we need to recall that, delta V(t)/delta V(0) = e^t/tau. Suppose that we want to know how long we have to wait until delta V(t)/delta V(0) = e^t/tau becomes 1/20. We'll call that time t_1/20 and have delta V(t_1/20)/delta V(0) = 1/20. This will allow us to solve for t_1/20? Find t_1/20 in terms of tau. Will t_1/20 be changed if the time constant for the circuit is increased? If it changes, explain how. [2] The choice of the fraction delta V9t)/delta V(0) is arbitrary. We could choose it to be 1/2 and usually do when dealing with radioactive decay. If we make this choice then the time for the voltage to fall to half of its initial value is called the half-life (and is written t_1/2). There is a relation between half-life and time constant that can be shown using delta V(t)/delta V(0) = e^t/tau. If t = t_1/2 then delta V(t_1/2)/delta V(0) = 1/2. Now use delta V(t)/delta V(0) = e^t/tau to show that t_1/2 = tau ln2. In the previous diagram. half the period of the generator (during which a single voltage decay occurs) is about 4 tau so the period of square waves from the signal generator is T approximately equal to 8 tau. We'd get a better idea of the voltage being asymptotically approached if we used square waves of a longer period. How long? We want the voltage across the capacitor to be 'small' enough. Recall delta V(t)/delta V(0) = e^t/tau. What is the ratio of the voltage after t = 20 tau to the initial voltage across the capacitor? [2] The implication is that we should use a low frequency such as f_low = 1/20 tau at which to measure time constants.

Answer 1

The time constant of an RC circuit is given by $\tau =RC$ Hence if capacitance is fixed, the time constant can be varied by varying thr resistance value.

While discharging the capacitor, The voltage V is given by $V(t)=V_{0}(e^{-t/\tau })$

Hence V(0) =V₀. $\therefore \frac{V(t)}{V_{0}}=e^{-t/\tau }$ . After a long time, t=infinity, V(infinity)=0 $\therefore \frac{V(t)}{V_{0}}=0$ .

t_1/20 says that the capacitor is discharged 1/20 of its initial voltage.

Hence $V_{0}/20=V_{0}(e^{-t^{1/20}/\tau })$ On solving this, we get,

$\tau ln(20)=t_{1/20}$ Yes if time constant increases, t_1/20 increases by a factor of ln(20)=3

Similarly, for t_1/2 , $V_{0}/2=V_{0}(e^{-t^{1/2}/\tau })$ . Hence on rearranging this like in the above expression, we get $\tau ln(2)=t_{1/2}$ .

$V(t)=V_{0}(e^{-t/\tau })$ at t>=5 time constants, the value of voltage approaches almost zero. Hence at t=20 time constants, The voltage across the capacitor is decayed mostlly and can be approximated to zero. Hence the ratio will also be zero.