Relativistic Momentum

Lheiman07 — Fri, 30 Oct 2026 05:36:39 GMT

Power

Lheiman07 — Fri, 30 Oct 2026 05:22:57 GMT

Difficult

@@ Line 14: / Line 14: @@
 :<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>
 where '''F''' is force, '''Δr''' is displacement, '''Δt''' is the duration of time and '''W''' is work.
 It then follows that instantaneous power is
-:<math>power = F\cdot v</math>
+:<math>power = \vec{F}\cdot\vec{v}</math>
 where '''v''' is velocity.
-The SI unit for power is watts (J/s)
+The SI unit for power is watts (J/s). The units can also be expressed in terms of fundamental SI units:
 ===A Computational Model===
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 ===Difficult===
 ==Connectedness==

User:Lheiman07

Lheiman07 — Fri, 30 Oct 2026 02:53:09 GMT

User account Lheiman07 was created

RC Circuit

Tommyg — Mon, 26 Oct 2026 13:57:04 GMT

Add Thevenin method and worked switched RC example

← Older revision		Revision as of 08:57, 26 October 2026
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	[[File:rccurrent.gif\|300px\|center\|]]		[[File:rccurrent.gif\|300px\|center\|]]

			== Analyzing a Switched Circuit with One Capacitor ==

			The basic series RC circuit is a useful starting point, but many circuit problems have several resistors and a switch. A reliable way to analyze any linear circuit with one capacitor is to replace the rest of the circuit, as seen from the capacitor’s terminals, with its Thevenin equivalent. The resulting circuit has one voltage source, one resistance, and the capacitor.

			This method works when the circuit contains one capacitor and resistors with constant values. It also works if several capacitors can first be combined into one equivalent capacitor. If there are multiple independent capacitors, the circuit may need more than one differential equation.

			=== The General Voltage Equation ===

			Suppose a switch changes position at time <math>t_0</math>. Define:

			* <math>V_i</math> as the voltage across the capacitor immediately before the switch changes.
			* <math>V_f</math> as the final, steady-state capacitor voltage after the switch changes.
			* <math>R_{\text{th}}</math> as the Thevenin resistance seen from the capacitor’s terminals after the switch changes.
			* <math>C</math> as the capacitance.

			For times after the switch changes, the capacitor voltage is

			[math]\displaystyle{ V_C(t)=V_f+(V_i-V_f)e^{-(t-t_0)/(R_{\text{th}}C)} }[/math]

			The time constant is

			[math]\displaystyle{ \tau=R_{\text{th}}C }[/math]

			This equation describes both charging and discharging. It also handles a capacitor that starts with a nonzero voltage and moves toward a different final voltage.

			The capacitor’s voltage cannot change instantly. Therefore,

			[math]\displaystyle{ V_C(t_0^+)=V_C(t_0^-)=V_i }[/math]

			The current can change instantly when the switch changes. If positive current is defined as entering the capacitor terminal labeled positive, then

			[math]\displaystyle{ I_C(t)=C\frac{dV_C}{dt}=\frac{V_f-V_i}{R_{\text{th}}}e^{-(t-t_0)/\tau} }[/math]

			A negative value means the actual current flows opposite to the chosen positive direction.

			=== How to Find the Thevenin Equivalent ===

			1. Draw the circuit as it is connected after the switch changes. The initial voltage <math>V_i</math> comes from the circuit before the switch changes; <math>R_{\text{th}}</math> and <math>V_f</math> come from the circuit afterward.
			2. Remove the capacitor and find the open-circuit voltage across its terminals. This is <math>V_f</math>, the voltage the capacitor approaches after a long time.
			3. Find <math>R_{\text{th}}</math> looking into the capacitor’s terminals. Turn off independent ideal voltage sources by replacing them with wires, and turn off independent ideal current sources by opening them. Leave dependent sources active.
			4. Calculate <math>\tau=R_{\text{th}}C</math>.
			5. Substitute <math>V_i</math>, <math>V_f</math>, and <math>\tau</math> into the general voltage equation.
			6. Use <math>Q=CV_C</math> or <math>I_C=C\,dV_C/dt</math> if the problem asks for charge or current.

			At one time constant, the voltage has moved about 63.2% of the way from its initial value toward its final value. At three time constants, it has moved about 95.0% of the way. At five time constants, it is more than 99% of the way there.

			=== Worked Example: A Capacitor Connected to a Voltage Divider ===

			A <math>12\ \text{V}</math> ideal battery is connected through a switch and a <math>2.0\ \text{k}\Omega</math> resistor, <math>R_1</math>, to a node. A <math>4.0\ \text{k}\Omega</math> resistor, <math>R_2</math>, and a <math>1000\ \mu\text{F}</math> capacitor are both connected from that node to ground. Before the switch closes, the capacitor is disconnected from the battery and has voltage <math>0\ \text{V}</math>. At <math>t=0</math>, the switch closes.

			Find the capacitor’s final voltage, the time constant, the voltage after one time constant, and its initial charging current.

			;Step 1: Find the initial voltage.
			The capacitor begins uncharged, so <math>V_i=0\ \text{V}</math>.

			;Step 2: Find the final voltage.
			After a long time, the capacitor current is zero. The resistors form a voltage divider, so

			[math]\displaystyle{ V_f=12\ \text{V}\left(\frac{4.0\ \text{k}\Omega}{2.0\ \text{k}\Omega+4.0\ \text{k}\Omega}\right)=8.0\ \text{V} }[/math]

			;Step 3: Find the Thevenin resistance.
			Turn off the ideal battery by replacing it with a wire. From the capacitor’s terminals, <math>R_1</math> and <math>R_2</math> are in parallel:

			[math]\displaystyle{ R_{\text{th}}=R_1\parallel R_2=\frac{R_1R_2}{R_1+R_2}=\frac{(2.0)(4.0)}{2.0+4.0}\ \text{k}\Omega=1.33\ \text{k}\Omega }[/math]

			;Step 4: Find the time constant.
			Since <math>1000\ \mu\text{F}=1.00\times10^{-3}\ \text{F}</math>,

			[math]\displaystyle{ \tau=R_{\text{th}}C=(1.33\times10^3\ \Omega)(1.00\times10^{-3}\ \text{F})=1.33\ \text{s} }[/math]

			;Step 5: Write the voltage as a function of time.
			For <math>t\geq0</math>,

			[math]\displaystyle{ V_C(t)=8.0\left(1-e^{-t/(1.33\ \text{s})}\right)\ \text{V} }[/math]

			At one time constant,

			[math]\displaystyle{ V_C(\tau)=8.0(1-e^{-1})\ \text{V}\approx5.06\ \text{V} }[/math]

			;Step 6: Find the initial current into the capacitor.
			Using the current direction into the capacitor’s positive terminal,

			[math]\displaystyle{ I_C(0^+)=\frac{V_f-V_i}{R_{\text{th}}}=\frac{8.0\ \text{V}}{1.33\times10^3\ \Omega}\approx6.0\ \text{mA} }[/math]

			The current decreases as the capacitor charges. After a long time, the capacitor current approaches zero, even though current still flows through the two resistors in the voltage divider.

			=== What to Check in a Circuit Sketch ===

			* Mark the capacitor’s positive terminal and use the same voltage direction throughout the calculation.
			* Use the circuit before switching to determine <math>V_i</math>. Capacitor voltage is continuous at the instant the switch changes.
			* Use the circuit after switching to find <math>V_f</math> and <math>R_{\text{th}}</math>.
			* At steady state in a DC circuit, an ideal capacitor carries no current. It acts like an open circuit after it has finished charging.
			* The resistance in <math>\tau=R_{\text{th}}C</math> is the resistance seen from the capacitor’s terminals. It may combine several resistors and is not always the resistor drawn next to the capacitor.
			* Use seconds for time, ohms for resistance, and farads for capacitance. Since <math>\Omega\cdot\text{F}=\text{s}</math>, the time constant has units of seconds.

			=== Practice Check ===

			A capacitor initially has voltage <math>6.0\ \text{V}</math>. After a switch changes position, the Thevenin equivalent seen by the capacitor is <math>10\ \text{V}</math> in series with <math>3.0\ \text{k}\Omega</math>. The capacitance is <math>200\ \mu\text{F}</math>.

			# Find the time constant.
			# Find the capacitor voltage after one time constant.
			# Find the initial current into the positive terminal.

			;Answers:

			[math]\displaystyle{ \tau=(3.0\times10^3\ \Omega)(200\times10^{-6}\ \text{F})=0.60\ \text{s} }[/math]

			[math]\displaystyle{ V_C(\tau)=10+(6.0-10)e^{-1}\approx8.53\ \text{V} }[/math]

			[math]\displaystyle{ I_C(0^+)=\frac{10-6.0}{3.0\times10^3}\approx1.33\ \text{mA} }[/math]

			=== Further Reading ===

			OpenStax, ''University Physics Volume 2'', Section 10.5, “RC Circuits”: https://openstax.org/books/university-physics-volume-2/pages/10-5-rc-circuits

	==Examples==		==Examples==

User:Tommyg

Tommyg — Mon, 26 Oct 2026 13:47:40 GMT

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Relativistic Momentum

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