Правка разделов «Statement», «Answer»

naz правка от 24 июн. 2026 г., 10:16

правка #19152 предыдущая #19151 ← раньше

en/11.2.10.md +9 −2

		@@ -1,7 +1,12 @@
			### Statement

		−	$11.2.10.$ ~~[Insert the problem statement]~~
		+	$11.2.10.$ The circuit of the electrical circuit includes a resistance $R$ and an uncharged capacitor of capacity $C$.

		+	a. Prove that the charge on the capacitor during the appearance and then disappearance of the magnetic flux through the circuit does not exceed the value of $\frac{ΦT}{CR^2}$, where $T$ is the lifetime of this magnetic flux, $Φ$ is its maximum value.
		+
		+	b**. To determine the direct current $I$ flowing in the circuit during the time $T$, if there is an alternating current caused by electromagnetic induction in the same time interval, the capacitance potential $V$ is measured after all currents disappear, and then the direct current is estimated using the formula $I =\frac{CV}{T}$.
		+	Determine the maximum error of such an estimate in the case when the amplitude of the alternating current is $k$ times greater than $I$.
		+
			### Solution

			Savchenko does not mention this in the condition, but clearly implies that $T << RC$ in both points of the problem.

			a) The derivation of equations for the charge on a capacitor as a function of time is not relevant to the topic of the chapter, therefore, with the reader's permission, it is not given here.

			So, let's take the equation for the charge on a capacitor as a function of time.:

			$$q = \varepsilon C (1 - e^{-\frac{t}{RC}})$$

			From here it is not difficult to derive the formula for the voltage across the capacitor:

			$$U = \varepsilon(1 - e^{-\frac{t}{RC}})$$

			Now let's think about it, these equations describe charging at constant EMF. This is clearly not our case, so $\varepsilon$ clearly depends on $t$. Moreover, the charge on the capacitor accumulates until $\varepsilon > U$. Therefore, in the equation for charge, in the extreme case, $\varepsilon = V$ should be substituted, and in the equation for voltage, the voltage that is induced by the field should be substituted.

			Let's use a fairly well-known limit for an argument tending to 0:

			$$e^x = 1 + x$$

			Let's rewrite our equations with this in mind, and don't forget to change the notation based on what was said above.:

			$$q = \frac{UT}{R}$$

			$$U = \frac{\varepsilon_{i} T}{RC}$$

			Obviously, we now need to maximize $U$. this means maximizing $\varepsilon_i$. Its maximum value:

			$$\varepsilon_i = \frac{\Phi}{T}$$

			From here:

			$$q_{max} = \frac{\Phi T}{CR^2}$$

			b) Despite the asterisk, using the previous paragraph, the solution will be quite simple.

			The condition means that the $RC$ circuit has a DC source and side noise $i(t)$ generated by the magnetic field. This means that the charge accumulated on the capacitor for a small $T$ is equal to:

			$$Q = IT + q_{ind}$$

			So the current strength through it is:

			$$\hat{I} = I + \frac{q_{ind}}{T}$$

			So there is a desired error.:

			$$\Delta I = \frac{q_{ind}}{T}$$

			But the maximum of this expression is achieved with the maximum charge that accumulates on the plates when the field is turned on for a short time, and we have already found this value. It only remains to make a reservation that in our expression from point a it is necessary to replace $\frac{\Phi}{R} = kIT$, because we are considering the worst case in which our DC pulse coincides in time with the peak value of the alternating current. Hence, taking into account the formula from point a, we get:

			$$\Delta I = \frac{kIT}{RC}$$
		@@ -56,4 +61,6 @@Solution

			#### Answer

		−	[Insert a concise answer or boxed result]
		+	а) Q.E.D.
		+
		+	b) $$\Delta I = \frac{kIT}{RC}$$

	@@ -1,7 +1,12 @@
	### Statement		### Statement

	$11.2.10.$ ~~[Insert the problem statement]~~		$11.2.10.$ The circuit of the electrical circuit includes a resistance $R$ and an uncharged capacitor of capacity $C$.

			a. Prove that the charge on the capacitor during the appearance and then disappearance of the magnetic flux through the circuit does not exceed the value of $\frac{ΦT}{CR^2}$, where $T$ is the lifetime of this magnetic flux, $Φ$ is its maximum value.

			b**. To determine the direct current $I$ flowing in the circuit during the time $T$, if there is an alternating current caused by electromagnetic induction in the same time interval, the capacitance potential $V$ is measured after all currents disappear, and then the direct current is estimated using the formula $I =\frac{CV}{T}$.
			Determine the maximum error of such an estimate in the case when the amplitude of the alternating current is $k$ times greater than $I$.

	### Solution		### Solution

	Savchenko does not mention this in the condition, but clearly implies that $T << RC$ in both points of the problem.		Savchenko does not mention this in the condition, but clearly implies that $T << RC$ in both points of the problem.

	a) The derivation of equations for the charge on a capacitor as a function of time is not relevant to the topic of the chapter, therefore, with the reader's permission, it is not given here.		a) The derivation of equations for the charge on a capacitor as a function of time is not relevant to the topic of the chapter, therefore, with the reader's permission, it is not given here.

	So, let's take the equation for the charge on a capacitor as a function of time.:		So, let's take the equation for the charge on a capacitor as a function of time.:

	$$q = \varepsilon C (1 - e^{-\frac{t}{RC}})$$		$$q = \varepsilon C (1 - e^{-\frac{t}{RC}})$$

	From here it is not difficult to derive the formula for the voltage across the capacitor:		From here it is not difficult to derive the formula for the voltage across the capacitor:

	$$U = \varepsilon(1 - e^{-\frac{t}{RC}})$$		$$U = \varepsilon(1 - e^{-\frac{t}{RC}})$$

	Now let's think about it, these equations describe charging at constant EMF. This is clearly not our case, so $\varepsilon$ clearly depends on $t$. Moreover, the charge on the capacitor accumulates until $\varepsilon > U$. Therefore, in the equation for charge, in the extreme case, $\varepsilon = V$ should be substituted, and in the equation for voltage, the voltage that is induced by the field should be substituted.		Now let's think about it, these equations describe charging at constant EMF. This is clearly not our case, so $\varepsilon$ clearly depends on $t$. Moreover, the charge on the capacitor accumulates until $\varepsilon > U$. Therefore, in the equation for charge, in the extreme case, $\varepsilon = V$ should be substituted, and in the equation for voltage, the voltage that is induced by the field should be substituted.

	Let's use a fairly well-known limit for an argument tending to 0:		Let's use a fairly well-known limit for an argument tending to 0:

	$$e^x = 1 + x$$		$$e^x = 1 + x$$

	Let's rewrite our equations with this in mind, and don't forget to change the notation based on what was said above.:		Let's rewrite our equations with this in mind, and don't forget to change the notation based on what was said above.:

	$$q = \frac{UT}{R}$$		$$q = \frac{UT}{R}$$

	$$U = \frac{\varepsilon_{i} T}{RC}$$		$$U = \frac{\varepsilon_{i} T}{RC}$$

	Obviously, we now need to maximize $U$. this means maximizing $\varepsilon_i$. Its maximum value:		Obviously, we now need to maximize $U$. this means maximizing $\varepsilon_i$. Its maximum value:

	$$\varepsilon_i = \frac{\Phi}{T}$$		$$\varepsilon_i = \frac{\Phi}{T}$$

	From here:		From here:

	$$q_{max} = \frac{\Phi T}{CR^2}$$		$$q_{max} = \frac{\Phi T}{CR^2}$$

	b) Despite the asterisk, using the previous paragraph, the solution will be quite simple.		b) Despite the asterisk, using the previous paragraph, the solution will be quite simple.

	The condition means that the $RC$ circuit has a DC source and side noise $i(t)$ generated by the magnetic field. This means that the charge accumulated on the capacitor for a small $T$ is equal to:		The condition means that the $RC$ circuit has a DC source and side noise $i(t)$ generated by the magnetic field. This means that the charge accumulated on the capacitor for a small $T$ is equal to:

	$$Q = IT + q_{ind}$$		$$Q = IT + q_{ind}$$

	So the current strength through it is:		So the current strength through it is:

	$$\hat{I} = I + \frac{q_{ind}}{T}$$		$$\hat{I} = I + \frac{q_{ind}}{T}$$

	So there is a desired error.:		So there is a desired error.:

	$$\Delta I = \frac{q_{ind}}{T}$$		$$\Delta I = \frac{q_{ind}}{T}$$

	But the maximum of this expression is achieved with the maximum charge that accumulates on the plates when the field is turned on for a short time, and we have already found this value. It only remains to make a reservation that in our expression from point a it is necessary to replace $\frac{\Phi}{R} = kIT$, because we are considering the worst case in which our DC pulse coincides in time with the peak value of the alternating current. Hence, taking into account the formula from point a, we get:		But the maximum of this expression is achieved with the maximum charge that accumulates on the plates when the field is turned on for a short time, and we have already found this value. It only remains to make a reservation that in our expression from point a it is necessary to replace $\frac{\Phi}{R} = kIT$, because we are considering the worst case in which our DC pulse coincides in time with the peak value of the alternating current. Hence, taking into account the formula from point a, we get:

	$$\Delta I = \frac{kIT}{RC}$$		$$\Delta I = \frac{kIT}{RC}$$
	@@ -56,4 +61,6 @@Solution

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		а) Q.E.D.

			b) $$\Delta I = \frac{kIT}{RC}$$