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What is the current in circuit with 6 v battery and 2 resistor?
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∙ 12y ago
3 ampere
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∙ 12y ago
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Two resistors are connected in series to a 9 volt battery. The larger resistor provides twice as many ohms resistance as the smaller one. The current through the circuit is 3 amps. What is the resista?
2
What is the voltage drop through a resistor of 6 ohms and 2 amps?
12 volts...! The voltage drop across a 2 ohm resistor depends on the current flowing through it. As voltage (E) equals current (I) times resistance (R), if 1/2 amp is flowing through your 2 ohm resistor, 1/2 times 2 = 1 volt. If 1 amp is flowing through your 2 ohm resistor, 1 times 2 = 2 amps. Piece of cake. If the two ohm resistor is the only component in the circuit, it will drop whatever the applied voltage is. Put a 2 ohm resistor across a 6 volt battery, it drops 6 volts. If you put your 2 ohm resistor across a 9 volt battery, it drops 9 volts. Another way to say voltage drop may help. The voltage drop across a resistor is the voltage it "feels" when in a circuit. And that last couple of examples says that very well. In a circuit where a given resistor is the only component, it drops all the voltage in the circuit. It "feels" all the voltage in the circuit. In a circuit where there are 2 resistors of equal value in series, each one drops or "feels" half of the applied voltage. (The sum of the voltage drops equals the applied voltage.) As you work more with simple circuits using resistors in different arrangements with a given voltage source, try thinking of the voltage drop of a resistor as the voltage it "feels" when the circuit is energized.
Is the heat loss and current of a resistor affected by being in a parallel circuit or can you just calculate it the same as in series?
The heat generated by any particular resistor depends (at least electrically) solely on the power it dissipates. Power dissipation in a resistor is equal to current squared times resistance, and the current through the resistor is equal to the voltage across it divided by the resistance. If we take a 10 ohm resistor ('your resistor') and put it in a series circuit such that there is 10 volts across your resistor, the current through it will be 1 ampere (10/10=1). the power dissipated will be 10 watts (1^2 * 10=10). If we put your resistor in a parallel circuit that also puts 10 volts across it, then the current and power will be the same. Your resistor does not know or care where the voltage came from. From this point of view, once you get down to the voltage across the resistor, it does not matter what type of circuit it is in. On the other hand, for any given power supply voltage, then the type of circuit and the value of external components certainly does affect the terminal voltage and thus the current through as well as the power dissipated by the resistor. In a parallel circuit, the voltage across your resistor remains basically the same no matter what resistance you put in parallel with it (unless you overload the power supply or the power supply has high internal resistance). In this case, the voltage across the resistor is the same as the power supply, current is I=E/R, R being that resistor only, and power is P=I^2 * R. In a series circuit the current through the resistors is I=E/R, R being the total resistance (including the other resistor(s)). The power dissipation in your resistor will then be P=I^2 * R, I being the series current we just calculated, and R being your resistor only. Since the other resistors affect the current, and since the current is the same no matter where you measure in a series circuit, then the voltage across your resistor and thus the power dissipation will be affected. The voltage across your resistor will be E=I*R, I being the series current we just calculated, and R being your resistor only. So, while the calculation for power dissipated in a particular resistor does not change relative to what type of circuit it is in, the calculation to arrive at the voltage across the resistor and/or the current through it (which you will then need to calculate power) does. Keep in mind there are other mechanical parameters that influence the actual case temperature of the resistor. Physical size of the case, composition, and airflow velocity, if any, will alter the case-to-ambient thermal conductivity. Ambient temperature will also be a factor in the final temperature.
A coil of 8 ohm parallel with coil of 20 ohm This is connected in series with 12 ohm resistor If the whole circuit is connected across a battery having emf of 30V internal resistance of 2ohm whats v?
v of what? v across what? v measured from what 2 points? v across the coils? v across the resistor? v across the coils and resistor? v across the battery? v across the battery and coils? v across the battery and resistor? or are you asking what v stands for? v stands for voltage.
A 5 ohm resistor a 20 ohm resistor and a 25 ohm resistor are all connected in series to a 100 volt power source What is the current running through the circuit?
To find the current running through the circuit, you need to use Ohm's Law, which states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them. You can use the following formula to calculate the current: I = V / R Where I is the current, V is the voltage, and R is the resistance. In this case, the total resistance of the circuit is the sum of the individual resistances, which is 5 ohms + 20 ohms + 25 ohms = 50 ohms. Therefore, the current flowing through the circuit is: I = 100 volts / 50 ohms = 2 amps.
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.
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.
What is the current in a 12-volt circuit with a 6-ohm resistor connected series?
2 A (amperes)
How a simple electrical circuit works?
This is most easily answered in a diagram but I will give it my best shot. The simplest electrical circuit consists of a battery, 2 conductors (wires), and a light. Electrons flow from the battery through the light and back to the battery. At this level of simplicity it is helpful to think of a bucket, 2 hoses, and a pump. To work like an electrical circuit, both hoses have to be full of water to begin with. When the pump is activated the water flows from the bucket, through the pump, and back to the bucket. Anything that interferes with the flow of water shuts off the water flow, called current in electrical circuits. ==== A simple circuit is a battery and a resistor. When they are connected, in the resistor current flows down a potential gradient, which produces heat.
Two resistors are connected in series to a 9 volt battery. The larger resistor provides twice as many ohms resistance as the smaller one. The current through the circuit is 3 amps. What is the resista?
2
What is the current in a 10V circuit if the resistance is 2Ω?
If a resistor has a current of 2 amperes when connected to a single battery, connecting a second identical battery in parallel with the first will not change the current through the resistor.This is because the voltage across the resistor will not change. Yes, the current capacity of the battery (pair) will double, but the voltage will not change.The exception is if the battery does not have the capacity to supply 2 amperes without sagging in output voltage. In this case, adding the second battery will slightly increase the current through the resistor.(By the way, without some kind of equalizing circuit, it is not a good idea to connect batteries in parallel. This is because slight differences in voltage could create substantial current flows through the batteries, and possibly damage, fire, and/or explosion.)
What is the voltage drop through a resistor of 6 ohms and 2 amps?
12 volts...! The voltage drop across a 2 ohm resistor depends on the current flowing through it. As voltage (E) equals current (I) times resistance (R), if 1/2 amp is flowing through your 2 ohm resistor, 1/2 times 2 = 1 volt. If 1 amp is flowing through your 2 ohm resistor, 1 times 2 = 2 amps. Piece of cake. If the two ohm resistor is the only component in the circuit, it will drop whatever the applied voltage is. Put a 2 ohm resistor across a 6 volt battery, it drops 6 volts. If you put your 2 ohm resistor across a 9 volt battery, it drops 9 volts. Another way to say voltage drop may help. The voltage drop across a resistor is the voltage it "feels" when in a circuit. And that last couple of examples says that very well. In a circuit where a given resistor is the only component, it drops all the voltage in the circuit. It "feels" all the voltage in the circuit. In a circuit where there are 2 resistors of equal value in series, each one drops or "feels" half of the applied voltage. (The sum of the voltage drops equals the applied voltage.) As you work more with simple circuits using resistors in different arrangements with a given voltage source, try thinking of the voltage drop of a resistor as the voltage it "feels" when the circuit is energized.
A 5 ohm resistor a 20 ohm resistor and a 25 ohm resistor are all connected in series to a 100 volt power source The current running through the circuit is?
E/R=I. 100/50=2 amps.
How much amount of heat will produce in 22kilo ohm resistor?
If the resistor is conducting electrical current, then the power it dissipates (heat energy per second) is(current through it)2 times (22,000)or(voltage across it)2 divided by (22,000).If the resistor is connected in an unpowered circuit, or stored in a drawer, then it dissipates zero heat.
Is the heat loss and current of a resistor affected by being in a parallel circuit or can you just calculate it the same as in series?
The heat generated by any particular resistor depends (at least electrically) solely on the power it dissipates. Power dissipation in a resistor is equal to current squared times resistance, and the current through the resistor is equal to the voltage across it divided by the resistance. If we take a 10 ohm resistor ('your resistor') and put it in a series circuit such that there is 10 volts across your resistor, the current through it will be 1 ampere (10/10=1). the power dissipated will be 10 watts (1^2 * 10=10). If we put your resistor in a parallel circuit that also puts 10 volts across it, then the current and power will be the same. Your resistor does not know or care where the voltage came from. From this point of view, once you get down to the voltage across the resistor, it does not matter what type of circuit it is in. On the other hand, for any given power supply voltage, then the type of circuit and the value of external components certainly does affect the terminal voltage and thus the current through as well as the power dissipated by the resistor. In a parallel circuit, the voltage across your resistor remains basically the same no matter what resistance you put in parallel with it (unless you overload the power supply or the power supply has high internal resistance). In this case, the voltage across the resistor is the same as the power supply, current is I=E/R, R being that resistor only, and power is P=I^2 * R. In a series circuit the current through the resistors is I=E/R, R being the total resistance (including the other resistor(s)). The power dissipation in your resistor will then be P=I^2 * R, I being the series current we just calculated, and R being your resistor only. Since the other resistors affect the current, and since the current is the same no matter where you measure in a series circuit, then the voltage across your resistor and thus the power dissipation will be affected. The voltage across your resistor will be E=I*R, I being the series current we just calculated, and R being your resistor only. So, while the calculation for power dissipated in a particular resistor does not change relative to what type of circuit it is in, the calculation to arrive at the voltage across the resistor and/or the current through it (which you will then need to calculate power) does. Keep in mind there are other mechanical parameters that influence the actual case temperature of the resistor. Physical size of the case, composition, and airflow velocity, if any, will alter the case-to-ambient thermal conductivity. Ambient temperature will also be a factor in the final temperature.
You are just asking that let us assume you have connected 2 or 3 resistors in a circuit and the current flowing through the circuit and from all the resistors will be same but how?
If the resistors are connected in series, the total resistance will be the sum of the resistances of each resistor, and the current flow will be the same thru all of them. if the resistors are connected in parallel, then the current thru each resistor would depend on the resistance of that resistor, the total resistance would be the inverse of the sum of the inverses of the resistance of each resistor. Total current would depend on the voltage and the total resistance
What is the current of each resistor and the total current drawn from the battery of a 5 ohm resistor 10 ohm resistor and a 15 ohm resistor in parallel connected to a 15 volt battery?
The resistors are 5, 10, and 15 ohms.The current through the 5-ohm resistor is E/R = 15/5 = 3 Amp.The current through the 10-ohm resistor is E/R = 15/10 = 1.5 Amp.The current through the 15-ohm resistor is E/R = 15/15 = 1 Amp.Their total effective resistance in parallel is the reciprocal of [ (1/5) + (1/10) + (1/15) ] =the reciprocal of [ (6/30) + (3/30) + (2/30) ] = the reciprocal of [ (11/30) ] = 30/11 ohms .The total current drawn from the battery is E/R = (15)/(30/11) = (15 x 11/30) = 11/2 = 5.5 Amp.Note:The 5-ohm resistor is dissipating 45 watts.The 10-ohm resistor is dissipating 22.5 watts.The 15-ohm resistor is dissipating 15 watts.The poor battery is delivering 82.5 watts.None of this is going to last very long at all.Most likely, the battery has already expired,and/or the 5-ohm resistor has already exploded,while we've been here playing with our calculators.
