When a transistor radio is switched off, the current falls away according to the differential equation #(dI)/dt=-kI# where #k# Is a constant . If the current drops to 10% in the first second ,how long will it take to drop to 0.1% of its original value?

Answer 1

# 3 \ s#

We have a current, #I#at time #t# in a circuit that flows according to the DE:
# (dI)/(dt) = -kI #

It is a First Order Separable Ordinary Differential Equation, allowing us to "separate the variables" and gather terms to obtain:

# int \ 1/I \ dI = int \ -k \ dt #

which are composed of standard integrals, allowing us to integrate directly to obtain:

# ln |I| = -kt + C #
# :. |I| = e^(-kt + C) #

Furthermore, since the exponential is positive across the board, we can write:

# I(t) = e^(-kt)e^(C) #
# \ \ = Ae^(-kt) #, say, where #A=e^(C)#
So, the initial current, #I(0)#, flowing (at time #t=0#) is:
# I(0) = Ae^(0) = A #

Since we know that in the first second the current decreases to 10% of its starting value, we can calculate:

# I(1) = Ae^(-k) #

And:

# I_1 = 10/100 * I(0) => Ae^(-k) = 1/10 * A #
# :. e^(-k) = 1/10 => k = ln(10) #

As a result, we can write the answer as:

# I(t) = Ae^(-tln10) #
We want the time, #T#, such that. #I(T)# is #0.1%# of the initial current #I(0)#, so we seek #T# satisfying:
# I(T) = 0.1/100 * I(0) => Ae^(-Tln10) = 0.1/100 * A #
# :. e^(-Tln10) = 1/1000 #
# :. -Tln10 = ln (1/1000) #
# :. Tln10 = 3ln10 #
# :. T= 3 #
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Answer 2

The given differential equation is ( \frac{dI}{dt} = -kI ).

We are told that when ( t = 1 ), ( I = 0.1I_0 ), where ( I_0 ) is the original value of the current.

Using separation of variables and integrating both sides, we get:

[ \int \frac{1}{I} , dI = \int -k , dt ]

This simplifies to:

[ \ln|I| = -kt + C ]

Using the initial condition when ( t = 1 ) and ( I = 0.1I_0 ), we find the constant ( C ):

[ \ln|0.1I_0| = -k + C ]

[ C = \ln|0.1I_0| + k ]

Now, substitute ( C ) back into the equation:

[ \ln|I| = -kt + \ln|0.1I_0| + k ]

[ \ln\left|\frac{I}{0.1I_0}\right| = -kt + k ]

[ \ln\left|\frac{I}{0.1I_0}\right| = k(-t + 1) ]

[ \frac{I}{0.1I_0} = e^{k(-t + 1)} ]

[ I = 0.1I_0 e^{k(1 - t)} ]

To find when the current drops to ( 0.001I_0 ), set ( I = 0.001I_0 ):

[ 0.001I_0 = 0.1I_0 e^{k(1 - t)} ]

[ 0.001 = 0.1e^{k(1 - t)} ]

[ 0.01 = e^{k(1 - t)} ]

Taking the natural logarithm of both sides:

[ \ln(0.01) = k(1 - t) ]

[ -4.605 = k(1 - t) ]

[ -4.605/k = 1 - t ]

[ t = 1 - \frac{4.605}{k} ]

This gives the time it takes for the current to drop to ( 0.1% ) of its original value.

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Answer 3

To find the time it takes for the current to drop to 0.1% of its original value, we need to solve the given differential equation (\frac{dI}{dt} = -kI).

Given that the current drops to 10% of its original value in the first second, we can use this information to find the value of (k).

Let's denote the original current as (I_0). After one second, the current is (0.1I_0). Using the given differential equation, we can set up the following equation:

[\frac{dI}{dt} = -kI] [\int_{I_0}^{0.1I_0} \frac{1}{I} dI = \int_{0}^{1} -k dt] [\ln\left(\frac{0.1I_0}{I_0}\right) = -kt \Bigg|_0^1] [\ln(0.1) = -k(1) - (-k(0))] [\ln(0.1) = -k] [k = -\ln(0.1)]

Now, we can use the value of (k) to find the time it takes for the current to drop to 0.1% of its original value.

Let (t) represent this time. We want to find (t) when (I = 0.001I_0). Substituting into the differential equation:

[\frac{dI}{dt} = -kI] [\int_{I_0}^{0.001I_0} \frac{1}{I} dI = \int_{0}^{t} -k dt] [\ln\left(\frac{0.001I_0}{I_0}\right) = -k t] [\ln(0.001) = -k t] [t = \frac{\ln(0.001)}{-k}]

Substitute the value of (k) we found earlier:

[t = \frac{\ln(0.001)}{-(-\ln(0.1))}]

[t = \frac{\ln(0.001)}{\ln(0.1)}]

[t \approx \frac{-6.9078}{-2.3026}]

[t \approx 3]

So, it will take approximately 3 seconds for the current to drop to 0.1% of its original value.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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