Question about Audio Players & Recorders

Ohms law can be applied here . When applying Ohms circuits are taken as perfect with no additional deviations to values introduced by connections or temperature, so although not exactly correct it is a very good guide. Almost all electical calcs can be done with Ohms law and its variations.

so

V(volts) = I(current) x R (resistance)

I x R =5000

so V =5000

Answer is 5000 Volts

Posted on Apr 30, 2008

Hi,

a 6ya expert can help you resolve that issue over the phone in a minute or two.

best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.

the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).

click here to download the app (for users in the US for now) and get all the help you need.

goodluck!

Posted on Jan 02, 2017

a transistor has 3 leads,[emitter current, collector current, base current, i am going to give you list of options that the transistors can be used for and it depends on the configuration of the circuits.

input resistance, current gain, voltage gain, output resistance, power gain, voltage gain, (input and output power) + [input and output voltage] and {input and output resistance} + input and output power gain.

18 various formulas to a range in groups of 2's

input resistance, current gain, voltage gain, output resistance, power gain, voltage gain, (input and output power) + [input and output voltage] and {input and output resistance} + input and output power gain.

18 various formulas to a range in groups of 2's

Apr 14, 2014 | Volfenhag ZX-7170 Car Audio Amplifier

All cars, made. EFI

1: fuse good.?

2: ECU sees key on, some cars run pump for 3 seconds now, many don't, read FSM under fuel pump it's all there.

3: The ECU energizes the Fuel pump relay as you crank engine.

the relay closes, and sends 12vdc to the pump.

so we check for 12v a the relay output cranking,

if dead, it can be ( bad Relay or ECU, usually never ECU)

bad relay or shorted output.?

be aware that pumps do short.

some pumps draw too much current, 2x normal(bad) and over load the supple line, (fuse didnt blow) this can cause the ECU or

other part to act wrong, (low voltage)

so we measure pump current first. to be sure its not too much.

if the voltage is 0v to pump , i unplug the pump.

crank got 12v? then the pump must be shorted.

or the relay is burned up.

keep in mind the voltage only matters if current is flowing.

most pumps draw about 5 amps.

if the draw is 5amp and voltage is 12v and the pump is dead.

its a bad pump.

if the draw is 10amps (overload) the pump is bad or wires cut and hitting body.

if the draw is 0 amps at 12v, then the wires are or the pump is dead.

this is how i diagnose all electrics.

this circuit is easy, its only 3 parts not counting wires. and fuse.

the ECU

the relay,

the pump.

btw, did you know i can find a soon to fail pump (works now)

using a scope/?

here is that magic. Duane here, is using current probe.

the pump need full power at all times to work correctly

a pump can actually have open commutator segments or or shorted windings (and still spin) the scope finds these very common electr. faults, with NO TAKING TANK FROM CAR !

http://www.realfixesrealfast.com/realfixesrealfast.com/Postal_L_L_V_Videos/Pages/LLV_Fuel_pump_drops_out.html

1: fuse good.?

2: ECU sees key on, some cars run pump for 3 seconds now, many don't, read FSM under fuel pump it's all there.

3: The ECU energizes the Fuel pump relay as you crank engine.

the relay closes, and sends 12vdc to the pump.

so we check for 12v a the relay output cranking,

if dead, it can be ( bad Relay or ECU, usually never ECU)

bad relay or shorted output.?

be aware that pumps do short.

some pumps draw too much current, 2x normal(bad) and over load the supple line, (fuse didnt blow) this can cause the ECU or

other part to act wrong, (low voltage)

so we measure pump current first. to be sure its not too much.

if the voltage is 0v to pump , i unplug the pump.

crank got 12v? then the pump must be shorted.

or the relay is burned up.

keep in mind the voltage only matters if current is flowing.

most pumps draw about 5 amps.

if the draw is 5amp and voltage is 12v and the pump is dead.

its a bad pump.

if the draw is 10amps (overload) the pump is bad or wires cut and hitting body.

if the draw is 0 amps at 12v, then the wires are or the pump is dead.

this is how i diagnose all electrics.

this circuit is easy, its only 3 parts not counting wires. and fuse.

the ECU

the relay,

the pump.

btw, did you know i can find a soon to fail pump (works now)

using a scope/?

here is that magic. Duane here, is using current probe.

the pump need full power at all times to work correctly

a pump can actually have open commutator segments or or shorted windings (and still spin) the scope finds these very common electr. faults, with NO TAKING TANK FROM CAR !

http://www.realfixesrealfast.com/realfixesrealfast.com/Postal_L_L_V_Videos/Pages/LLV_Fuel_pump_drops_out.html

Nov 09, 2013 | Previa Resister, fuel pump. SWITCH &...

(Voltage) = (Current) x (Resistance)

so

1. Voltage is 200 Volts

2. Resistance is 44 Ohms

so

1. Voltage is 200 Volts

2. Resistance is 44 Ohms

Sep 02, 2013 | Office Equipment & Supplies

Which current are you referring to ??......

If you think of an electrical device as a piece of plumbing, voltage is the amount of water that you send down into the pipe, resistance is the pipe's relative width or narrowness, and current is the speed with which the water flows.

Power measures the water's relative difficulty or ease making its way through the pipe.

You relate all these values to one another using a common set of physics equations known as Ohm's law.

If you need to calculate electricity's current flow, you'll need to have at least two of the three values -- voltage, resistance or power -- listed above.

Calculate current flow using voltage and resistance.

According to Ohm's law, you can express electricity's current in amps as a ratio of its voltage in volts to the resistance of the device it's flowing through in ohms -- I = E/R, respectively.

For example, if you want to know the current flow of 220 V of electricity as it flows through a laptop computer with 80 ohms of resistance, you would simply plug these values into the equation as follows: I = 220/80 = 2.75 amps.

Calculate current flow using power and resistance.

Ohm's law also states that electrical current, "I," is equal to the square root of the power dissipated as it travels through the device divided by that device's resistance.

If a light bulb dissipates 80 watts of power and has a resistance of 55 ohms, you can calculate the electricity's current as follows: I = sqrt(80/55) = sqrt(1.4545) = 1.20 amps.

Calculate current flow using power and voltage.

If you have a space heater which dissipates 420 watts of power when it takes in 120 V of electricity, Ohm's law states you can calculate this electricity's current using the equation "I = P/E." For this example, compute current like so: I =420/120 = 3.5 amps.

http://www.the12volt.com/ohm/ohmslaw.asp

If you think of an electrical device as a piece of plumbing, voltage is the amount of water that you send down into the pipe, resistance is the pipe's relative width or narrowness, and current is the speed with which the water flows.

Power measures the water's relative difficulty or ease making its way through the pipe.

You relate all these values to one another using a common set of physics equations known as Ohm's law.

If you need to calculate electricity's current flow, you'll need to have at least two of the three values -- voltage, resistance or power -- listed above.

Calculate current flow using voltage and resistance.

According to Ohm's law, you can express electricity's current in amps as a ratio of its voltage in volts to the resistance of the device it's flowing through in ohms -- I = E/R, respectively.

For example, if you want to know the current flow of 220 V of electricity as it flows through a laptop computer with 80 ohms of resistance, you would simply plug these values into the equation as follows: I = 220/80 = 2.75 amps.

Calculate current flow using power and resistance.

Ohm's law also states that electrical current, "I," is equal to the square root of the power dissipated as it travels through the device divided by that device's resistance.

If a light bulb dissipates 80 watts of power and has a resistance of 55 ohms, you can calculate the electricity's current as follows: I = sqrt(80/55) = sqrt(1.4545) = 1.20 amps.

Calculate current flow using power and voltage.

If you have a space heater which dissipates 420 watts of power when it takes in 120 V of electricity, Ohm's law states you can calculate this electricity's current using the equation "I = P/E." For this example, compute current like so: I =420/120 = 3.5 amps.

http://www.the12volt.com/ohm/ohmslaw.asp

Aug 13, 2013 | Computers & Internet

All meters capable of resistance measurements require batteries to power the circuit under test and meter movement.

Some manufacturers prefer to use 2 different voltage sources to be applied to specific ranges offered. They use the lower voltage for low resistance ranges and higher voltages for high resistance ranges. Doing this help prevent larger currents (and wattages) from flowing in the component or circuit under test when the low resistance ranges are selected.

Higher resistance components and circuits will have significantly lower current flow (and lower wattages), so a higher voltage is supported and provides greater resolution of measured values on the meter.

I hope this helps & good luck!

Some manufacturers prefer to use 2 different voltage sources to be applied to specific ranges offered. They use the lower voltage for low resistance ranges and higher voltages for high resistance ranges. Doing this help prevent larger currents (and wattages) from flowing in the component or circuit under test when the low resistance ranges are selected.

Higher resistance components and circuits will have significantly lower current flow (and lower wattages), so a higher voltage is supported and provides greater resolution of measured values on the meter.

I hope this helps & good luck!

Oct 05, 2012 | Measuring Tools & Sensors

typically you can charge or discharge large capacitors using a resistor. Not just any resistor will do. You will need one with high-enough resistance and wattage ratings or you'll burn up the resistor. I would use a 1/2W, 4.7kohm or a 1W, 1kohm. These are minimum values and you can use a slightly higher resistance but it will take longer to charge/discharge.

To charge, connect the positive terminal, then hold the resistor between the ground wire and the ground terminal for 30-60 seconds. Then connect the ground terminal. To discharge, disconnect both terminals, then hold the resistor across the terminals (do not reconnect terminals without charging). In either case, you can use your fingers to hold the resistor as DC will not shock you, but if the resistor gets hot it will burn you. I use needle-nose pliers to hold the resistor (just in case). Just be careful not to short the resistor leads or terminals with the pliers.

Explanation: For a standard cap 1 - 2 farads, the discharge voltage is up to 20V so we'll figure off of that. I would use at least a 1/2 watt resistor, a 1 watt could be safer, depending on the value. For a 1 watt, the current can be no more than .05A at 20V. For a 1/2 watt, only .025A. Let's say .04A and .02A, respectively. Ohms = Volts / Amps, so 20V / .04A gives us 500ohms. Double that to be safe (avoid overheating) and it's 1kohm. Same process for the 1/2W version will be 2kohm, but it's not as hefty, so I would recommend 4.7kohm.

To charge, connect the positive terminal, then hold the resistor between the ground wire and the ground terminal for 30-60 seconds. Then connect the ground terminal. To discharge, disconnect both terminals, then hold the resistor across the terminals (do not reconnect terminals without charging). In either case, you can use your fingers to hold the resistor as DC will not shock you, but if the resistor gets hot it will burn you. I use needle-nose pliers to hold the resistor (just in case). Just be careful not to short the resistor leads or terminals with the pliers.

Explanation: For a standard cap 1 - 2 farads, the discharge voltage is up to 20V so we'll figure off of that. I would use at least a 1/2 watt resistor, a 1 watt could be safer, depending on the value. For a 1 watt, the current can be no more than .05A at 20V. For a 1/2 watt, only .025A. Let's say .04A and .02A, respectively. Ohms = Volts / Amps, so 20V / .04A gives us 500ohms. Double that to be safe (avoid overheating) and it's 1kohm. Same process for the 1/2W version will be 2kohm, but it's not as hefty, so I would recommend 4.7kohm.

Jun 29, 2010 | Lanzar 16 Farad Power Capacitor LQ16CAP

Try reading the resistance on the coil itself. If your coil is damaged and has a high resistance the voltage will still read 120 volts, becuase there is no current. The voltage drop across the coil is equal to the resistance of the coil times the current. V = I * R. "I" being current. If the resistance of the coil is high because it is broken then the voltage will still read 120V. Another way is to read the current through the coil.

If it is a 1500W fryer then the current through the coil is about 12 amps. 1500/120= 12.5. To calculate the resistance of the coil you would you take the voltage divided by the current. 120/12.5 = 9.6 ohms.

Therefore the resistance of the coil should be very low. It will actually be lower when it is not hot. So your ohmeter should read somewhere between 5 and 15 ohms.

Bob Janelli

If it is a 1500W fryer then the current through the coil is about 12 amps. 1500/120= 12.5. To calculate the resistance of the coil you would you take the voltage divided by the current. 120/12.5 = 9.6 ohms.

Therefore the resistance of the coil should be very low. It will actually be lower when it is not hot. So your ohmeter should read somewhere between 5 and 15 ohms.

Bob Janelli

Oct 01, 2009 | Kitchen Appliances - Others

The voltage difference can be overcome. The problem will be the difference in frequency 50hz vs 60hz. The motors would have to be replaced with motors designed to operate at 50hz. The electronics would most likely need to be replace as well.

Best bet would be to contact whirlpool international to see if the offer a conversion kit/package.

Best bet would be to contact whirlpool international to see if the offer a conversion kit/package.

Feb 02, 2009 | Whirlpool LER5636P Electric Dryer

graydog,

if you manage to change the keypad most probably the condenser MIC inside the phone been misaligned. try to check if the condeser MIC is properly sitted on it's original place and if you do have a multi tester you might as well check the resistance of the condenser MIC.

the nromal resistance for condeser MIC is 500ohms and some are 60 omhs.

cheers

mark

if you manage to change the keypad most probably the condenser MIC inside the phone been misaligned. try to check if the condeser MIC is properly sitted on it's original place and if you do have a multi tester you might as well check the resistance of the condenser MIC.

the nromal resistance for condeser MIC is 500ohms and some are 60 omhs.

cheers

mark

Jan 17, 2009 | Motorola Mobility V262

69 people viewed this question

Usually answered in minutes!

×