RC and RL time constant and the time to charge or discharge to any percent; series RLC resonant frequency, impedance, phase, Q factor and bandwidth.
अपडेट किया गया जाँचे गए उदाहरण: 7
τ = 1 s: after one τ the capacitor voltage has risen to 63.2% of 5 V, and it takes 2.303 s to reach 90%. After 5τ it is within 1% of the final value.
An RC circuit responds with time constant τ = RC and an RL circuit with τ = L/R: after one τ the capacitor voltage or inductor current has covered 63.2% of the way to its final value, and reaching a fraction p takes t = −τ ln(1 − p). For a series RLC circuit the calculator finds the resonant frequency f₀ = 1/(2π√(LC)), the impedance and phase at your signal frequency from the reactances 2πfL and 1/(2πfC), and Q = (1/R)√(L/C).
The default, 10 kΩ with 100 µF on a 5 V supply, gives τ = 1 s and 2.30 s to reach 90%. RC time constants set timer delays, switch debounce periods and filter cut-off frequencies, 1/(2πRC). With 10 Ω, 10 mH and 1 µF in series, f₀ = 1,591.55 Hz and Q = 10.
Components are ideal and the input is a step: no capacitor leakage and no inductor winding resistance beyond R. After 5τ the response is within 0.67% of its final value (e⁻⁵), the usual rule for "fully charged".
जाँच का स्रोत: Python 3.8 math: τ = 10⁴ × 10⁻⁴ = 1 s; t = τ ln 10; E = ½CV² = ½·10⁻⁴·25
जाँच का स्रोत: Python 3.8 math: 5(1 − e⁻¹) = 3.1606028
जाँच का स्रोत: Python 3.8 math: ln 2 = 0.6931472 (τ = 1 s)
जाँच का स्रोत: Python 3.8 math: τ = L/R = 10⁻⁴ s, I = 5/100, i(50 µs) = 0.05(1 − e^−0.5) = 0.0196735
τ = R × C, in seconds when R is in ohms and C in farads. It is the time a charging capacitor takes to reach 63.2% (1 − e⁻¹) of the supply voltage, or a discharging one to fall to 36.8%. 10 kΩ with 100 µF gives τ = 1 s; 1 kΩ with 1 µF gives 1 ms.
About five time constants. After 5τ the voltage is 99.33% of the supply (1 − e⁻⁵), which engineers treat as fully charged; in theory it never quite gets there. Reaching 90% takes τ ln 10 ≈ 2.303τ and 99% takes τ ln 100 ≈ 4.605τ, so a 10 kΩ, 100 µF circuit reaches 99% in 4.6 s.
f₀ = 1/(2π√(LC)), with L in henries and C in farads; in a series circuit R does not shift it. 10 mH with 1 µF resonates at 1,591.55 Hz, and 100 µH with 100 pF at 1.59 MHz. At f₀ the inductive and capacitive reactances are equal and cancel, so the impedance falls to R alone and the current peaks.
Q = (1/R)√(L/C), the ratio of either reactance at resonance to the resistance. It sets the −3 dB bandwidth, Δf = f₀/Q: 10 Ω, 10 mH and 1 µF give Q = 10 and a 159 Hz band around 1,592 Hz. Above Q = 0.5 the circuit rings (underdamped); exactly 0.5 is critically damped, reached here at R = 2√(L/C) = 200 Ω.
τ = L/R. It is the time for the current to rise to 63.2% of its final value V/R after the switch closes, or to fall to 36.8% once the source is removed and the coil discharges through R. 10 mH with 100 Ω gives τ = 100 µs, so on 5 V the current reaches 99% of 50 mA in about 460 µs.
सटीकता आपके इनपुट और विधि की मान्यताओं पर निर्भर करती है। दशमलव गणना 50 सार्थक अंकों का उपयोग करती है, लेकिन अनुमान, संख्यात्मक विधियाँ और स्रोत डेटा कम सटीक हो सकते हैं। दिखाए गए मानों को पूर्णांकित करने से ये सीमाएँ दूर नहीं होतीं। स्वतंत्र स्रोतों के हल किए गए उदाहरणों से जाँच: 7। उदाहरण के लिए, “10 kΩ and 100 µF charging to 90%” की जाँच Python 3.8 math: τ = 10⁴ × 10⁻⁴ = 1 s; t = τ ln 10; E = ½CV² = ½·10⁻⁴·25 से की गई है।
OpenStax University Physics Volume 2, §10.5 RC circuits; §14.4 RL circuits; §15.3 RLC series circuits with AC; HyperPhysics — Series RLC resonance and Q.
इस कैलकुलेटर में स्वतंत्र स्रोतों के उत्तरों वाले 7 हल किए गए उदाहरण शामिल हैं। ये परीक्षण समूह में चलते हैं और आप इन्हें यहाँ भी चला सकते हैं।
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