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Two equations for power are: P=I2R and P=EI where P=power, E=voltage, and I=current.
Solving the first equation for I and the second equation for E gives: I=(P/R)1/2 E=P/I
Now substitute the first equation into the second.
E=P/(P/R)1/2 = (P1/2)(R1/2)
Using the given values:
E = (5001/2)(51/2) = ((5)(100))1/2(51/2) = (51/2)(10)(51/2) = 50 volts
ANSWER: WHATEVER THE SOURCE VOLTAGE MINUS THE IR DROP OF WIRES
What else can I help you with?
A cell of emf 3.0 volts and internal resistance of 0.01 ohm is connected through an ammeter of 0.05-ohm internal resistance to a 5.0 rheostat by wire having a total resistance of 0.85-ohm?
total voltage = 4.5V, total resistance = 3.5 ohms, loop current = 4.5V / 3.5 ohms = 1.286Atotal voltage = 9V, total resistance = 4 ohms, loop current = 9V / 4 ohms = 2.25Atotal voltage = 13.5V, total resistance = 4.5 ohms, loop current = 13.5V / 4.5 ohms = 3Aetc.There is no solution to your problem conditions.
What voltage would be necessary to produce a current of 3A if the resistance of the curcuit were 50 OHMS?
The voltage is gained by multiplying the current and resistance together, i.e.. 50 x 500 = 25000 Imagine the three as a triangle with the voltage at the top, and the current and resistance at the bottom- V . ---- . I x R The voltage divided by the current is the resistance and the voltage divided by the resistance is the current. Therefore the current times the resistance is equal to the voltage. Having any two of these figures allows you to find the third.
What will happen to the current and voltage of a 120V supply with internal resistance of 0.1 ohms as each bulb having a resistance of 5 ohm is added across it in parallel?
Number of bulbs . . . Load voltage . . . Load current 0 . . . . . . . . . . 120 V . . . . . . . . 0 A. 1 . . . . . . . . . . 117.65 V .. . . . . 23.53 A. 2 . . . . . . . . . . 115.38 . . . . . . . 46.15 3 . . . . . . . . . . 113.21 . . . . . . . 67.92 4 . . . . . . . . . . 111.11 . . . . . . . 88.88 5 . . . . . . . . . . 109.09 . . . . . . . 109.09
What benefit is there for getting higher open circuit voltage?
There is no particular benefit for having a higher open-circuit (or 'no-load') voltage. In fact, an ideal voltage source would have no internal resistance and, therefore, its open-circuit voltage would be identical to its closed-circuit voltage.
How does voltage remains constant in resistors connected in parallel?
It is not necessary for anything about them to be the same.All that is necessary is that they create the desired equivalent resistor, having the correct resistance and power dissipation rating without overstressing either resistor.
Why an ideal voltimer must infinity resistor?
An ideal voltmeter must have infinite resistance to prevent it from drawing any current from the circuit it is measuring. If it had finite resistance, it would alter the voltage across the component being measured, leading to inaccurate readings. By having infinite resistance, the voltmeter ensures that it does not influence the circuit, allowing for precise voltage measurements.
What happens when you have low resistance?
Having low resistance means that there is an easier flow of electric current through a circuit. This can lead to higher current levels which could potentially damage the components in the circuit or cause overheating. It can also result in a decrease in voltage across the circuit.
How does the voltage divider biasing network got its name?
Because that's what it is. It is a voltage divider. Two resistors in series will have the same current through them. This is Kirchoff's current law. By Ohm's law, then, the voltage across each resistor is current times resistance, and this is a linear function. By Kirchoff's voltage law, then, the total voltage drop across both resistors is equal to the input voltage. Two resistors of the same value will divide the voltage in half. One resistor (the top resistor) having twice the resistance of the other, will divide the voltage to a third. If the top resistor is nine times the bottom resistor, the divider produces one tenth. And so on and so forth...
A cell of emf 3.0 volts and internal resistance of 0.01 ohm is connected through an ammeter of 0.05-ohm internal resistance to a 5.0 rheostat by wire having a total resistance of 0.85-ohm?
total voltage = 4.5V, total resistance = 3.5 ohms, loop current = 4.5V / 3.5 ohms = 1.286Atotal voltage = 9V, total resistance = 4 ohms, loop current = 9V / 4 ohms = 2.25Atotal voltage = 13.5V, total resistance = 4.5 ohms, loop current = 13.5V / 4.5 ohms = 3Aetc.There is no solution to your problem conditions.
What will happen if a voltmeter has a low shunt resistance instead of having high shunt resistance?
The purpose of a voltmeter is to indicate the potential difference between two points in a circuit.When a voltmeter is connected across a circuit, it shunts the circuit. If the voltmeter has a low resistance,it will draw a substantial amount of current. This action lowers the effective resistance of the circuit andchanges the voltage reading.
What voltage would be necessary to produce a current of 3A if the resistance of the curcuit were 50 OHMS?
The voltage is gained by multiplying the current and resistance together, i.e.. 50 x 500 = 25000 Imagine the three as a triangle with the voltage at the top, and the current and resistance at the bottom- V . ---- . I x R The voltage divided by the current is the resistance and the voltage divided by the resistance is the current. Therefore the current times the resistance is equal to the voltage. Having any two of these figures allows you to find the third.
Is fet having negative resistance?
JFET BFW20 shows negetive resistance when gate is grounded (VGS = 0) and vary Drain to source voltage and measure Drain current. As the voltage is increased, the drain current decreases. Prof.S.Lakshminarayana.
What will happen to the current and voltage of a 120V supply with internal resistance of 0.1 ohms as each bulb having a resistance of 5 ohm is added across it in parallel?
Number of bulbs . . . Load voltage . . . Load current 0 . . . . . . . . . . 120 V . . . . . . . . 0 A. 1 . . . . . . . . . . 117.65 V .. . . . . 23.53 A. 2 . . . . . . . . . . 115.38 . . . . . . . 46.15 3 . . . . . . . . . . 113.21 . . . . . . . 67.92 4 . . . . . . . . . . 111.11 . . . . . . . 88.88 5 . . . . . . . . . . 109.09 . . . . . . . 109.09
What benefit is there for getting higher open circuit voltage?
There is no particular benefit for having a higher open-circuit (or 'no-load') voltage. In fact, an ideal voltage source would have no internal resistance and, therefore, its open-circuit voltage would be identical to its closed-circuit voltage.
What is the effect caused by ammeter resistance when an ammeter is inserted into a circuit to measure the current?
The effect the multimeter might have on the circuit when inserted to measure the current is to increase the circuit resistance and decrease the available voltage to the circuit. This is because the multimeter in amps or milliamps mode does have a small resistance which is not zero, so by Ohm's law, there is a voltage drop across the multimeter; small, but not zero. Usually this effect is small. One way to compensate is to start by measuring voltage, and then inserting a separate ammeter and adjusting the power supply to match the original voltage. Of course, the voltmeter must be downstream of the ammeter.
Why series resistor is necessary when diode is forward bias?
THINK! (famous sign used in IBM for decades to motivate employees) What would happen if an ideal voltage source (i.e. no internal resistance) was connected directly across a diode in the forward bias direction? For a real silicon diode the knee voltage is about 0.7V and the diode's internal resistance can drop roughly another 0.3V before overloading. If the voltage source applied more than the sum of the knee voltage and the maximum voltage the internal resistance can drop before overloading (this sum is roughly 1V) then bad things will happen. You should have been able to figure out you answer yourself from the above. The situation above is only trivially changed by the use of a real voltage source (i.e. having internal resistance), it only postpones the overload a little.
How can you connected two heating elements each having 1000 watts to get lowest heating?
Connect the two heaters in series. The supply voltage, what ever the value, will be split in half across the two heaters. If the voltage is dropped by half across the two heaters in series, the wattage of each heater will be dropped to one quarter of the heaters wattage rating at its full voltage rating. The new wattage's will be added together for a new total wattage for the circuit. When you find the heaters working voltage use these two formulas; This one to find the resistance of the heater at its working voltage value, R = E (squared)/Watts. Then to find the watts at the reduced voltage value, W = E (squared)/R. R being the heaters resistance in ohms from the first formula.
