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I'm having some trouble focusing in on the schematic of the circuit.
Wiki User
∙ 12y ago
Wiki User
∙ 12y agoUse Ohm's law. C = V/R
C = current
V = voltage
R = resistance
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How are current and voltage affected in series and parallel circuits?
Voltage is Current multiplied by Resistance. The formulas for calculating the resistance in parallel and series circuits are: Resistors in Series: R(total)=R1+R2+R3+... Resistors in Parallel: 1/(Rtotal)=(1/R1)+(1/R2)+... Current is a measure of Coulombs of charge per unit of time or I=C/t The current depends on the circuit, whether it has capacitors or resistors, and the exact layout. Current 'flows' through wires in much the same way that water flows through pipes, so if the current meets some resistance (a resistor), some of the current will go through the resistor, but the rest will go through any other available path (like in a parallel circuit).
What is series connection?
You have two resistors, each with resistance of 12Ω, and a 12-volt battery. 1). The resistors are in series across the battery. ..... A. voltage across each resistor ..... B. current through each resistor ..... C. power dissipated by each resistor ..... D. total power delivered by the battery 2). The resistors are in parallel across the battery. ..... A. voltage across each resistor ..... B. current through each resistor ..... C. power dissipated by each resistor ..... D. total power delivered by the battery ============================================ 1). ... A. 6 volts ... B. 0.5 Amp ... C. 3 watts ... D. 6 watts 2). ... A. 12 volts ... B. 1 Amp ... C. 12 watts ... D. 24 watts
When a resistor is connected to a 25.0 V power supply the current through the resistor is 12.5 A.Calculate the resistance of a second resistor connected in series that reduces the current to 5.00 A.?
R2 = 3 ohms Explanation: For a circuit you can use ohms law where: V=I*R Where V is the voltage difference throughout the surface, I is the current, and R is the total resistance of the circuit. In your case you want to find the resistance so you have to change the formula to: R=V/I R of first circuit = 25volts/12.5amps = 2 ohms R of second circuit= 25 volts / 5 amps = 5 ohms The resistors here are connected in series which means that the resistance of the two can be added together. This gives you: Rtot= R1+R2 we found R of the first resistor by calculating the resistance in the first circuit. We also found Rtot which is the resistance in the second circuit, when you connect the two resistors together in series. Rtot=2 ohms+R2 5ohm=2ohms+R2 R2 = 3 ohms If the resistors where connected in parallel you cannot simply add the resistance. In that case: (1/Rtot)=(1/R1)+(1/R2) Hope that helps
What is a unit used to measure resistance?
The " Ohm " is. 1 ohm is the resistance across which 1 volt of EMF appears when the current through it is 1 Ampere.
What is the effect on current through a circuit of steady resistance when the voltage is doubled?
The current will increase and will flow more. If voltage increases, current must increase.
How current passes through resistor.?
What is the current running through resistor four?1 amps..!What is the current running through resistor one? 3 amps...!What is the current running through resistor three? 2amps..!What is the current running through resistor five? 3 amps..!What is the voltage drop running through resistor five? 45 volts...!What is the equivalent resistance through the parallel portion of the circuit? 6 ohmsAnswerA resistor is a conductor, albeit one with a higher resistance than a length of wire, so current passes through it without any problem. The magnitude of the current will, of course, be somewhat lower because of the additional resistance.
What is the resistance of a resistor if the potential differennce across the resistor is 9 V when current of 1 A flows through the resistor?
9 ohms
A 120 ohm resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source What is the current running through the 120 ohm resister?
If they're in parallel, then the resistors have no effect on each other. The current through each one is the same as it would be if the others were not there at all. The current through the 120Ω resistor is 120 volts/120Ω = 1 Ampere. The 60Ω and the 40Ω are red herring resistors.
If two bulbs in a series are connect to a battery and the current in the first bulb is found to be 1 ampere then what is the current through the battery?
A circuit with a 2 ohm resistor and a 4 ohm resistor in series with a 12 volt battery will have 2 amps flowing through each resistor. The current is the same in each resistor because they are in series, and a series circuit has constant current throughout.
What is the voltage drop running through resistor 1 resistor 1 equals 3 ohms?
There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
The voltage drop across a resistor is 1.0 V for a current of 3.0 A in the resistor what is the current that will produce a voltage drop of 9.0 V across the resistor?
The resistor is 1/3 of an ohm. A 9 volt drop across the resistor would cause a draw of 27 amps through the resistor. The wattage you would need for that resistor is at least a 243 watts.
What is direct current through resistor of 1 k-Ohm connected with12 volt battery?
12 milliamps
How can you convert a resistor of 270.4 watts to ohms without having been given voltage or current?
There is no way to make the conversion. The wattage given as that of the resistor is only the amount of power it is dissipating. It's the heat the resistor is radiating. The resistor is said to be using or radiating 270.4 watts. (That's a lot of wattage! How hot would a 270 watt incandescent lamp get? Very hot.) A watt is sometimes called a volt-amp. That's because watts equals volts times amps. And it's easy to see how that works with an example including the approximately 270 watts set down in the question. As regards that approximately 270 watts specified, if the resistor was a 1 ohm resistor and had 270 amps of current flowing through it, the 270 watts would be the wattage the resistor would be radiating. If the resistor was a 270 ohm resistor and had 1 amp of current flowing through it, it would (also) be radiating 270 watts. See how that works? An unlimited number of variations on the theme exist. To find the resistance of the resistor, one of two things must be known in addition to the wattage the resistor is running at. Either the voltage across the resistor or the current flow through the resistor must be specified to "finish" the problem.
Resistors A and B are connected in parallel If resistor B has twice the resistance of A what is the current taken by resistor B?
half of the current flowing thru resistor 1.... V=IR.
When an inductor and resistor are connected in parallel by how many degrees is the current flow through the resistor out of phase with the current flow through the inductor?
When a resistor and an inductor are both connected to an AC supply, the current in the resistor is in phase with the voltage, while the current in the inductor is a quarter-cycle (90 degrees) behind. Supposing they both draw 1 amp on a 12-volt AC supply. The resistor will dissipate 12 watts, while the inductor will dissipate no power. Any power that enters the inductor comes back to the generator in a later part of the cycle. But the current drawn from the supply is 1.414 amps, so this would be a load with a power factor of 0.707.
How do you measure current without an amp meter?
Depends on the current. Put a resistor in-line with the current, then measure the voltage across the resistor. V=RI. So, divide the measured voltage by resistor value. Be careful with the size of the resistor, as Power dissipated in a resistor is R*I^2 or V^2/2. So, a 1-Amp current into a 1 Ohm resistor will result in a 1Watt power dissipated in the resistor. If it's too small, it'll burn. Also, notice that if you do that, you haven't measured the current in the original circuit. You've measured the current when an extra resistor is installed in the original circuit, and that's different.
When you say your increasing the range of an ammeter what exactly are you doing?
You're connecting a resistance in parallel with the ammeter, so that when the meter and the additional resistor are inserted into a current-carrying circuit, a part of the current goes through the meter, and the rest of the current bypasses the meter and goes through the resistor instead. If you know exactly how much the split is, then you can calculate the larger total current from the smaller number that you read on the meter, so the meter is now able to measure and display more current than it could before ... the range of its measurement ability has been extended. Example: -- You know the resistance of the meter. -- You connect a resistor with exactly that same resistance in parallel with the meter. -- Now 1/2 of the current goes through the meter and 1/2 goes through the resistor. -- The total current is exactly double what you read on the meter.
