Hot answers tagged

17

No. The circuits may be on different phases or on opposing split phase. If one plug is removed from its socket its pins will be live due to backfeed from the other circuit. This presents an electrocution hazard. You'll have two parallel circuit breakers if they are on the same phase. Don't mess around with this. Install the proper circuit.


16

A load resistor is actually a bit of an abstract term... If you consider an electrical circuit is intended to act upon some other device in order to perform "work" then that external device is the "LOAD" of the circuit. simulate this circuit – Schematic created using CircuitLab However it is not that simple, since load must have a reference. ...


13

But what does it mean when one says,"It is a capacitive load" It means the load behaves like a capacitor. You have to deliver charge to it before the voltage across it will change. why output impedance should be low for capacitive loads? Because if you want to change the voltage across this load, you'll have to deliver a significant current. You ...


13

It may be worth slogging through the exact math to illustrate what's happening here. Assuming a source resistor \$R_S\$ and a load resistor \$R_L\$ the current through both resistors will simply be $$I = V\frac{1}{R_S+R_L}$$ We can calculate the voltage over the load as a simple voltage divider $$V_L = V\frac{R_L}{R_S+R_L}$$ The power over the load becomes $$...


12

A 'load resistor' is simply a resistor that is being used as a load. It may be very small, or may need to be physically big, depending on how much power it has to dissipate. When you see descriptions of circuits, various resistors might be qualified by what they do, so you may see resistors called 'feedback', 'damping', 'source', 'bias', 'potential divider'...


9

A load resistor is well... nothing but a resistor : a 2-terminal component that complies with Ohm's law and whose impedance is real (purely resistive, no reactance of admittance whatsoever). What makes it a load resistor is the fact that it is placed at the output of something. The key here is understanding that, actually, a load resistor (or a resistive ...


8

Will a capacitor bank and high current relay do it? Ignoring the inductive load for the moment since that raises an entirely different question, simply assuming that you need to supply 20A @ 12V for 5 seconds, the energy required is: $$12V \cdot 20A \cdot 5s = 1200J$$ Let's do a quick calculation to get a feel for the size of capacitor bank to deliver ...


8

From an electrical perspective loading refers to what is impeding powerflow. "weak pullup", refers to a high ohmage resistor. Likewise with generation and other such powersources a low load is one that requires low power flow & this implies higher impedance. Loading and impedance are inverse terms. Low load --> High resistance High load --> Low ...


8

This is just a conceptual model, and won't hold up to much scrutiny, but might help get your mind going in the right direction. Imagine your circuit is a piece of tubing. The power source is you blowing into one end. DC could be you blowing with a steady pressure, while AC could be you blowing in and out. Now, cap the other end and put a tiny hole in it......


8

If you under stand energy transfer, then power transfer is the same thing. Energy is \$E=P*t\$ where \$P\$ is in watts and \$t\$ is in seconds. Power transfer is just instantaneous energy transfer. Take a simple circuit: simulate this circuit – Schematic created using CircuitLab Now recall Ohm's law and the power law: $$ V = IR $$ $$ P = VI $$ If \$...


7

Inductance in the primary of a transformer decreases as the load on the secondary increases. No it doesn't. It may seem like it does (because when loaded your transformer takes more current into the primary) but just imagine that the load you put on the secondary (say 1:1) ratio were applied to the primary - the current in the load would be the same (1:1 ...


7

Using the water analogy, a 10 mm pipe will only allow a certain amount of water to pass, and it will depend on the applied pressure. If you double the pressure, the flow will double. A certain material will require a certain voltage to flow a certain current. Quoting a nice answer in Physics.SE about gravitation: Physics [Ohm's law] does not answer ...


7

You need to put a load on the battery to see if it has any charge left. Without a load, it may show an acceptable voltage, but when you actually try to use it the voltage drops because the battery is nearly dead. So to see if a battery is really usable you must measure the voltage when the battery is connected to a load. Like this: Dead Battery, no load, ...


7

Do the math. We don't know what your load is, other than it takes 20 W at 12 V. Its current at 12 V is therefore (20 W)/(12 V) = 1.67 A. That means, at least at the 12 V operating point, its effective equivalent resistance is (12 V)/(1.67 A) = 7.2 Ω. Without knowing more about the load we can't really know what its equivalent resistance at other ...


7

It's just a normal resistor. It's called a load resistor because it is there to add a load to the circuit. There is an implication that it will be dissipating a reasonable amount of power (otherwise it wouldn't be much of a load) but this isn't a requirement. e.g. early linear regulators required a minimum load to ensure voltage regulation, you would ...


6

No, the page is corrrect. The voltage drop across the resistor is equal to the current through the resistor multiplied by the resistance. V=IR The current is zero, therefore the voltage drop is zero, and Vs is present at the output.


6

Typical creep (parameter "1") of a Vishay load cell over the stated 30 minute interval. The Vishay document states: OIML recommendation R76 requires a 30 minute test and specifies an error limit for this time period, as well as the last 10 minutes (20 to 30 minutes). NTEP requires a one hour test and specifies an error limit for this period. OIML ...


6

The only way to know the actual self-discharge rate of any given battery/cell, is by measuring from a fully charge state to a discharged state, over a given time period. Aside from that, a look at the datasheet, or an educated guess based on battery chemistry is required. Standard NiMH Rechargeables for example, have a high discharge rate, noted of up to 4% ...


6

Cp is the input capacitance plus stray capacitance. You can use a few pF (3-5pF) for the value unless something is really strange. So, for a crystal rated with a 10pF load, Cl = (10pF-Cp)\$\cdot\$2, so if we use 4pF for Cp, we get 12pF for the load capacitors. For the 16pF crystal, Cl = (16-Cp)\$\cdot\$2, so using 4pF for Cp we get 24pF, using 3pF we get ...


6

A capacitance \$C\$ has impedance $$Z_{C} = \frac{1}{j\omega C}$$ But note that $$\frac{1}{j} = -j$$ so equivalently $$Z_{C} = -j\frac{1}{\omega C}$$ A capacitor's impedance therefore has a negative imaginary part. An inductor \$L\$ has impedance $$Z_{L} = j\omega L$$ and therefore has a positive imaginary part. So if the load has a positive ...


6

At any given frequency the network has an impedance that is just a single (complex) number, and if the imaginary part is negative then it's capacitive (at that frequency) and if the imaginary part is positive then it's inductive (at that frequency). If the real part is much bigger than the imaginary part, then it behaves mostly like a resistor. It's not ...


6

You can rearrange the circuit: simulate this circuit – Schematic created using CircuitLab But using Thévenin equivalence, you have this: simulate this circuit Which is a standard RC circuit with $$ V'_{in} = {R_2\over{R_1+R_2}}V_{in} $$ and $$ R_p = R_1||R_2 = {R_1R_2\over{R_1+R_2}}$$


6

Let's imagine your example circuit, except with the inductor replaced with a resistor. This is a really simple circuit: when the transistor is on, the current through it is given by V = IR. When the transistor is off, no current flows. Now, add the inductor back in. The inductor does not like the current flowing through it to change: however many amps it is ...


6

The LED does not light up because the B-E junction of the transistor is passing enough current to pull the 6V source down to <1V, which is too low to light the LED. It's amazing that the transistor doesn't burn out in this situation, but perhaps it's a rather beefy one, and your 6V supply has a relatively high impedance.


6

You've identified the problem fairly accurately, for capacitative load the instantaneous current requirements are quite large. In many designs short circuit detection simply disconnects a load when a short is detected reconnecting it when the short is gone . In this case your capacitor does not ever fully charge before the short circuit detection triggers, ...


6

I would use an easier approach: the capacitor is only needed to act as buffer for the high-current spikes. Just get the duration of the (biggest) spike (t_duration) and the height of the (biggest) spike (I_max) from the data you got ("For the time period of interest, I have the load's current draw over time."). The capacitor's voltage should never drop below ...


5

The real question is why would you think that the two capacitors need to be the same, which implies that they serve the same purpose? Often we do see them the same. That's the coincidence you should be asking about. There seeems to be a lot of religious beliefs and so-called "rules" out there for crystal capacitors, but none of those are substitutes for ...


5

Isn't there be a voltage drop across the resistor Rs?? If not, why not?? No, there is zero volts across a resistor, by Ohm's Law, when there is zero current through the resistor. If the load is an open circuit, there is zero current through \$R_S\$ and so, the voltage across \$R_S\$ is \$0V\$. By KVL: $$V_S = V_{Rs} + V_{OUT} = 0V + V_{OUT}$$ and so, ...


5

I found this document (Schematic Checklist), from the chip manufacturer (Microchip) which states (note point 5). Usually something like a high-value bias resistor is supplied internally for a Pierce oscillator configuration typically used on IC crystal oscillators, but some ICs don't have them, and perhaps this chip either does not have one, or the ones on ...


5

Sometimes it's easier to think of a resistor in terms of conductance. For instance, a 1 ohm resistor has a conductance of 1 amp per volt - if you put 5 volts across it the current will be 5 amps. A 100 ohm resistor has a conductance of 0.01 amps per volt etc.. Conductance is the reciprocal of resistance.


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