Low Frequency Electronic Transformer
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- Q: A real inductor can be modeled as an ideal inductor in series with an internal resistance as illustrated in the figure above.A time-varying current I is passed through a real inductor. At a time when I +4 A and dI/dt +14 A/s, the voltage drop across the inductor is observed to be Va - Vb +185 V. At another time when I +4 A and dI/dt -14 A/s, the observed voltage drop is V'a- V'b +43 V.
- Ldi/dt +4i Vab, L(14) + R(4) 185 L(-14) + R(4) 43 System of linear eqns, add together; 8R 228, R 28.5 ohm
- Q: A circuit is composed of a 35.0 ohm resistor and a 0.250 Henry inductor in series, powered by an AC source with frequency of 30.0 Hz. The resulting peak current is 0.100 A. a) What is the peak voltage of the resistor?b) What is the peak voltage of the inductor?c) What is the peak voltage of the power source?Please explain how to solve this to me, Thanks!!
- a) Resistor peak voltage is VRp Ip * R 0.1A * 35 ohm 3.5 V b) Inductor reactance at f 30Hz is XL 2pi*30Hz*0.25H 47.1 ohm Inductor peak voltage is VLp Ip * XL 0.1A * 47.1 ohm 4.71 V c) Total series circuit impedance is Z sqrt(R^2 + XL^2) 58.7 ohm Source peak voltage is Vp Ip * Z 0.1A * 58.7 ohm 5.87 V
- Q: An In depth answer would help me more
- i'm professional-existence, and my answer is no. The argument right it is fallacious. The organ donation already took place. This quest could have been phrased: professional-choicers: could you help a regulation permitting you to music down people you ve donationed organs to, and take those organs decrease back, all the mutually as ramming a suction tube into the decrease back of their head to suck their brains out purely to make greater advantageous confident that they die?
- Q: I understand that power is that rate at which work is done and that because of this the power is equal to d/dt (1/2Li^2). I also understand that the power is also equal to Li di/dt since Ldi/dt is v and v*i is power. I understand that since the power is equal to both of these equations that they are equal to each other. The part that I don't get is mathematically how to get from one to the other.
- you have written the steps yourself, from what i can tell d/dt [ (1/2)Li^2 ]. L is constant, i d/dt [ q(t) ], not constant using the power rule for differentiation d/dt [ (1/2)Li^2 ] (2)(1/2)Li*(di/dt). implicit differentiation if that step did not make sense, review the calculus techniques from first semester calc,, i dq/dt v -dΦ/dt -d/dt ( BAcosθ ) || B || (μ0 / 4π) * i visualize the charge (q) running through wire approaching inductor, as it flows into the coil, a B-field is induced, which tries to resist the CHANGE in current, this is where the work occurs
- Q: Hello,i need some help solving this one. i'm kind of lost of which formula to apply. Thanks so much for your help. Answer Key (B) A 45-mH inductor is connected to an ac source of emf with a frequency of 400 Hz and a maximum emf of 20 V. The maximum current is:A) 0B) .018 AC) 1.1 AD) 360 AE) 2300 A
- The reactance (XL) of an inductor is given by: XL 2πfL where f is the frequency in Hz and L is the inductance in henrys. XL 2π * 400 * 45/1000 XL 113 ohms Now it is just Ohm's Law i v / XL i 20 / 113 i 0.177 A i ≈ 0.18 A { I suspect a typo for answer (B) it should be 0.18 A }
- Q: A 265 mH inductor whose windings have a resistance of 6.70 Ω is connected across a 10.5 V battery with an internal resistance of 3.35 Ω. How much energy (in J) is stored in the inductor?Thanks for the help!
- U Potential energy R Resistance 6.7ohms I current V/R (10.5V/6.7ohms) 1.57A L inductance 265mH .265H U (.5)LI^2 (given formula in any physics text book) U (.5)*(.265H)*(1.57A)^2 .33 J
- Q: A 10.2*10^-6H inductor with negligible resistance and a resistor are the only elements in a circuit in which a current is flowing.What must be the value of the resistor so that the current will decrease to 50.0% of its maximum value in 1.10ms ?
- I Io e^(-t/τ) and τ L/R Given I/Io 0.5 e^(-1.1e-3/τ) Taking log to the base e and Solving we get τ L/R 0.00158690 R 10.2 /0.00158690 6428 ?
- Q: A 150 Omega resistor is connected in series with a 0.250-H inductor. The voltage across the resistor is v_R(3.80 V) *[cos(720*rad/ s)*t]Derive an expression for the circuit current. and Derive an expression for the voltage v_L across the inductor.I got by the equation I v/R.0253A*cos((720*rad/s)*t)and for the other i got by the equation v_LL* (di/dt) -(4.55*V)*sin((720 rad/s)*t)
- Not quite answering, but just pointing out that your sin and cos expressions could be written as (ex) sin[(720 rad/s) * t] Now when you substitute a value for t, the units inside the brackets become radians after cancellation and then all is well in the ups and downs. I think you have answered the question well, as long as your calculus is done correctly.
- Q: An RL circuit consists of two inductors of self-inductance L1 10.00 H and L2 4.00 H connected in parallel to each other, and connected in series to a resistor of 6.00 ? and a battery of 4.00 V (See figure). Assume that the inductors have no mutual inductance. When the battery is suddenly connected what is the inital rate of change of the current in inductor L1?Figure:
- Initially no current flows in L1 or L2, so the voltage drop across R 0 and thus 4 V is applied to L1 and L2. Use the formula V LdI/dt to find dI1/dt and dI2/dt. P.S. You did a bad copy of the imageshack page URL. What you posted actually ends with just as it appears, which is not a URL. That's because you used a simple copy command (ctrl-C or whatever) instead of Copy Shortcut. I'm assuming you have R + (L1 || L2), where + means 'in series with' and || means 'in parallel with'. EDIT: OK, got your URL and it's the way I assumed it was and fits your description.
- Q: An inductor and a resistor are in series with a sinusoidal voltage source. The frequency is set so that the inductive reactance is equal to the resistance. If the frequency is increased, then(A) VL 2VR(B)VL VR(C) VL VR(D)VR VL
- C. Inductive reactance increases with frequency. Z 2piFL.
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1. Manufacturer Overview
| Location | Shenzhen, Guangdong, China (Mainland) |
| Year Established | 2006 |
| Annual Output Value | US$2.5 Million - US$5 Million |
| Main Markets | North America; South America; Eastern Europe; Southeast Asia; Africa; Oceania; Mid East; Eastern Asia; Western Europe; Central America; Northern Europe; Southern Europe; South Asia; Domestic Market |
| Company Certifications | CE Certificates |
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3. Manufacturer Capability
| a) Trade Capacity | |
| Nearest Port | Shekou,Yantian |
| Export Percentage | 51% - 60% |
| No.of Employees in Trade Department | 3-5 People |
| Language Spoken: | English, Chinese |
| b) Factory Information | |
| Factory Size: | 3,000-5,000 square meters |
| No. of Production Lines | 9 |
| Contract Manufacturing | OEM Service Offered Design Service Offered Buyer Label Offered |
| Product Price Range | Average |
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Low Frequency Electronic Transformer
- Ref Price:
-
- Loading Port:
- China Main Port
- Payment Terms:
- TT or LC
- Min Order Qty:
- 800 Pieces pc
- Supply Capability:
- 10000 Pieces per Month pc/month
OKorder Service Pledge
Quality Product, Order Online Tracking, Timely Delivery
OKorder Financial Service
Credit Rating, Credit Services, Credit Purchasing
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