0
What else can I help you with?
An electric dc circuit is attached to a 12 volt battery and has a 3ohm resistor in it What is the current flowing through the circuit?
The current depends on the total effecvtive resistance of everything connectedacross the battery.If the resistor is the only component there, then the current is E/R = 12/3 = 4 amperes.
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.
What is a surge resistor?
A surge resistor is a resistor installed in a circuit to prevent a "surge" when conditions arise where a surge might occur. That might be a bit wordy, but that's the answer. Let's look at an instance where a surge resistor is used and see how it works. There are a number instances where energizing a circuit is followed by a surge of current. Like almost all the time. When that circuit "comes on" after the switch is flipped, current is flowing. And sometimes we need to place a resistor in series with the component or components that want to draw a ton of current on startup. We can look at a component that might just do that. Capacitors (caps) are sometimes fitted with something that limits a big shot of current when a circuit is energized. Caps, you recall, begin operation in the circuit of a modern equipment uncharged, and when these caps are first "seen" by the voltage source at the moment it's turned on, they look like a short circuit. Lots of current may want to flow. Jumping ahead, the cap sits in the operating circuit fat, dumb and happy "using" only a little current after things are going. But at startup, it needs to be checked to prevent it from "sucking up" (or sinking) too much current and overloading the supply. The surge resistor is on the job each time the circuit is energized. There are other applications for a surge resistor, but the purpose is the same - prevent "too much current" from flowing at a given moment so the circuit can transition into a "normal operating mode" (where current flow is more modest). The surge resistor is similar to a so-called current limiting (or, perhaps, ballast) resistor, but the name applied to the resistor in question might have more to do with the specific application.
Why must resistance never be measured in an active circuit?
First of all, in an active circuit, there are voltages flowing through various nodes. If you try to measure the resistor, those voltages will cause error in your meter, or may damage your meter. Second of all, even if you cut the power, any loop in the circuit around the resistor will affect the measured impedance. Unless you are certain that only the resistor will see the current introduced by the meter, you must take the resistor out of circuit to measure it.
Should a resistor go on the positive or negative side of an LED?
It does not matter. Kirchoff's Current Law states that the signed sum of the currents entering a node is zero. A consequence of that law is that the current in every part of a series circuit is the same. The only thing that resistor location affects is the potential voltage of the LED terminals with respect to the rest of the circuit. Certainly, if you are driving the LED with high voltage, such as 120VAC, you should consider the resistor location so as to reduce electrocution hazard but, the LED's performance is not affected by resistor location in the circuit.
If the resistance in the circuit is increased what will happen to the current and voltage?
* resistance increases voltage. Adding more resistance to a circuit will alter the circuit pathway(s) and that change will force a change in voltage, current or both. Adding resistance will affect circuit voltage and current differently depending on whether that resistance is added in series or parallel. (In the question asked, it was not specified.) For a series circuit with one or more resistors, adding resistance in series will reduce total current and will reduce the voltage drop across each existing resistor. (Less current through a resistor means less voltage drop across it.) Total voltage in the circuit will remain the same. (The rule being that the total applied voltage is said to be dropped or felt across the circuit as a whole.) And the sum of the voltage drops in a series circuit is equal to the applied voltage, of course. If resistance is added in parallel to a circuit with one existing circuit resistor, total current in the circuit will increase, and the voltage across the added resistor will be the same as it for the one existing resistor and will be equal to the applied voltage. (The rule being that if only one resistor is in a circuit, hooking another resistor in parallel will have no effect on the voltage drop across or current flow through that single original resistor.) Hooking another resistor across one resistor in a series circuit that has two or more existing resistors will result in an increase in total current in the circuit, an increase in the voltage drop across the other resistors in the circuit, and a decrease in the voltage drop across the resistor across which the newly added resistor has been connected. The newly added resistor will, of course, have the same voltage drop as the resistor across which it is connected.
An electric dc circuit is attached to a 12 volt battery and has a 3ohm resistor in it What is the current flowing through the circuit?
The current depends on the total effecvtive resistance of everything connectedacross the battery.If the resistor is the only component there, then the current is E/R = 12/3 = 4 amperes.
Do you need to apply a load to see a voltage drop across a resistor?
You'll see a voltage drop across a resistor if current is flowing through it. It only has to be a part of a complete circuit, i.e. one in which current is flowing.
What is the difference between a normal resistor and a bleeder resistor?
A resistor is a resistor. Plain and simple. By Ohm's Law, resistance in ohms is voltage in volts divided by current in amperes. The difference lies in application, not in the resistor itself. A normal resistor will introduce a voltage drop or current that makes some effect in the circuit, based on some design criteria. A bleeder resistor, on the other hand does not really affect the circuit - it is only there to "bleed off", or discharge, capacitors when the power is turned off. Consequently, a bleeder resistor will typically have a higher resistance than a normal resistor but, again, the issue is circuit design, not the resistor itself.
An electric DC circuit is attached to a 12 volt battery and has a 3 ohm resistor in it find the current flowing through the circuit?
If the 3-ohm resistor is the ONLY thing in the circuit, then the current flowing through it is (12 volts)/(3 ohms) = 4 amperes. If there are other things in the circuit besides the resistor, then the current depends on all of them.
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
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.
What does the resistor in a steady hand game do?
In a steady hand game, the resistor is part of an electrical circuit that detects when the player touches the wire with the metal loop. When the loop makes contact with the wire, it completes the circuit, allowing current to flow. The resistor limits the amount of current passing through, which helps prevent damage to the circuit and ensures that the buzzer or light indicator activates only when the circuit is completed improperly. This design enhances the gameplay by providing an immediate response to mistakes.
What is a surge resistor?
A surge resistor is a resistor installed in a circuit to prevent a "surge" when conditions arise where a surge might occur. That might be a bit wordy, but that's the answer. Let's look at an instance where a surge resistor is used and see how it works. There are a number instances where energizing a circuit is followed by a surge of current. Like almost all the time. When that circuit "comes on" after the switch is flipped, current is flowing. And sometimes we need to place a resistor in series with the component or components that want to draw a ton of current on startup. We can look at a component that might just do that. Capacitors (caps) are sometimes fitted with something that limits a big shot of current when a circuit is energized. Caps, you recall, begin operation in the circuit of a modern equipment uncharged, and when these caps are first "seen" by the voltage source at the moment it's turned on, they look like a short circuit. Lots of current may want to flow. Jumping ahead, the cap sits in the operating circuit fat, dumb and happy "using" only a little current after things are going. But at startup, it needs to be checked to prevent it from "sucking up" (or sinking) too much current and overloading the supply. The surge resistor is on the job each time the circuit is energized. There are other applications for a surge resistor, but the purpose is the same - prevent "too much current" from flowing at a given moment so the circuit can transition into a "normal operating mode" (where current flow is more modest). The surge resistor is similar to a so-called current limiting (or, perhaps, ballast) resistor, but the name applied to the resistor in question might have more to do with the specific application.
What happens if 3 resistors connected in parallel and one of the resistors is open circuited?
When one resistor in a parallel circuit is open-circuited, it effectively becomes an infinite resistance and no current flows through it. The total resistance of the parallel circuit increases, but the remaining resistors continue to function normally. The overall current through the circuit will decrease because the total current is now only dependent on the remaining active resistors. The voltage across all parallel resistors remains the same as before the open circuit occurred.
Why must resistance never be measured in an active circuit?
First of all, in an active circuit, there are voltages flowing through various nodes. If you try to measure the resistor, those voltages will cause error in your meter, or may damage your meter. Second of all, even if you cut the power, any loop in the circuit around the resistor will affect the measured impedance. Unless you are certain that only the resistor will see the current introduced by the meter, you must take the resistor out of circuit to measure it.
Should a resistor go on the positive or negative side of an LED?
It does not matter. Kirchoff's Current Law states that the signed sum of the currents entering a node is zero. A consequence of that law is that the current in every part of a series circuit is the same. The only thing that resistor location affects is the potential voltage of the LED terminals with respect to the rest of the circuit. Certainly, if you are driving the LED with high voltage, such as 120VAC, you should consider the resistor location so as to reduce electrocution hazard but, the LED's performance is not affected by resistor location in the circuit.
