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If two resistors have same voltage drop in a series circuit it means?
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∙ 6y ago
It means the two resistors have same resistance
Wiki User
∙ 6y ago
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What is the PD across two resistors connected parallel to a battery of emf 6V?
The potential difference across two resistors connected in parallel to a battery with a potential difference of 6 volts is 6 volts. Kirchoff's Voltage Law: The signed sum of the voltage drops in a series circuit is zero. This means that that the two series circuits involving the battery and each resistor have the same voltage across each other, and the series circuit involving the two resistors have the same voltage across each other.
When adding resistance in a series circuit what happens to supply voltage?
In a series circuit... Kirchoff's current law: The sum of the signed currents entering a node is zero. Since a series circuit consists of only nodes each connected to only two elements, this means that the current in every point in a series circuit is the same. Kirchoff's voltage law: The sum of the signed voltage drops in a series circuit is zero. This means, that if you segregate the sources from the loads, the total voltage across all the nodes is equal to the total voltage across all the sources. That may seem trite, but take the case where you have one battery in series with two resistors also in series. If you know the voltage across one resistor, then you know the voltage across the other resistor - it is the battery voltage minus the first resistor's voltage. Ohm's law: Voltage is current times resistance. This actually applies everywhere; series circuits, parallel circuits, DC circuits, AC circuits, etc.
2ohm and 4ohm resistor are connected in series to a 12V battery what is the current through each resistor and use Ohm's law 2 show that the voltage drops across the individual resistors add up to 12V?
To answer this question, you need not only Ohm's law, but also Kirchoff's current and voltage Laws.Kirchoff's current law say the current at every point in a series circuit is the same. Lets call that current i.Ohm's law say voltage is resistance times current. That means the voltage across each resistor R1 and R2 is V1 = R1i and V2 = R2i.Kirchoff's voltage law says that the signed voltage drops around a series circuit add op to zero. This means that the voltages V1 and V2 must add up to be be equal to 12. This means that R1i + R2i = 12. This means that i = 12/(R1 + R2). But, hey, you say, this is just Ohm's law! This means that the current through the circuit is 2A. It also means that the resistance of two resistors in series is simply the sum of their resistance.Now that we know the current, we can calculate each resistor's voltage. V1 = R1i = (2)(2) = 4, and V2 = R2i = (4)(2) = 8. Crosscheck 4 + 8 = 12, so the two voltages do add up to twelve volts, as expected.In summary: The current through both resistors is 2A. The voltage across the 2 ohm resistor is 4V. The voltage across the 4 ohm resistor s 8V.
What is the behavior of current and voltage in a series cricuit?
Kirchhoff's Current Law: The sum of the signed currents entering a node is zero. A consequence of this is that, in a series circuit every node only has two connections, one entering and one leaving, thus, in a series circuit, the current is the same at every point. Kirchhoff's Voltage Law: The sum of the signed voltage rises going around a series circuit is zero. This means, for example, that if you have two voltage drops, such as two bulbs in series with a battery, the voltage drops across them will add up to the voltage across the battery.
Which of kichhoff's laws applies to series circuits?
Kirchoff's Current Law states that the signed sum of the currents entering a node is equal to zero. In a simple parallel circuit, say with one battery and two light bulbs, this means the current coming out of the battery will be exactly equal to the sum of the currents entering the two light bulbs. In a series circuit, it also means that the current at every point in the circuit is the same. A parallel circuit can be construed as a special case of a series circuit, when you start to combine elements. Kirchoff's Voltage Law states that the signed sum of the voltage drops around a series circuit is equal to zero. Since a parallel circuit can be construed as a special case of a series circuit, this means that voltage across parallel nodes is equal.
What is the PD across two resistors connected parallel to a battery of emf 6V?
The potential difference across two resistors connected in parallel to a battery with a potential difference of 6 volts is 6 volts. Kirchoff's Voltage Law: The signed sum of the voltage drops in a series circuit is zero. This means that that the two series circuits involving the battery and each resistor have the same voltage across each other, and the series circuit involving the two resistors have the same voltage across each other.
If three 6 volt batteries are connected in series the overall voltage in circuit is?
Regardless of the number and value of the resistors, total voltage drop in a series circuit will equal the voltage rise, or the applied voltage. Apply 6 volts to three series resistors and the sum of the voltage drops will be 6 volts. No mystery here. Think it through and it will lock in. To get you ready for more "advanced" analysis, Kirchhoff said the algebraic sum of the voltages in any closed loop is zero. Going all the way around a series circuit, we'd encounter the battery, and all the series resistors. The battery is a voltage rise, and the resistors are voltage drops. The polarity of a voltage rise is opposite that of a voltage drop. This means that when they are added algebraically, if they are equal, they will sum to zero. Work this with a battery connected across a single resistor to get a handle on it. You'll need the ideas to manage calculations in loops of parallel circuits. Remember that in any closed loop, the algebraic sum of the voltages is zero.
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.
Definition of voltage divider?
Basically, if you have two resistors in series, then the total resistance is Rt = R1+R2. According to Kirchhoff's law, the total current entering a junction must be equal to the total current leaving it, so for a series circuit the current is the same in both resistors. From Ohm's law V=IRt and so from above V = I(R1+R2). This means that that total voltage in the circuit, V, is equal to the sum of the voltages across each resistor V1=IR1 and V2=IR2. This "divides" the voltage, so that for a 30V supply with two resistors of 10 ohms and 20 ohms respectively, the voltage across the first resistor will be 10V and the voltage across the second will be 20V. In this way a component requiring a lower voltage than the supply voltage can tap off from one of the resistors.
When adding resistance in a series circuit what happens to supply voltage?
In a series circuit... Kirchoff's current law: The sum of the signed currents entering a node is zero. Since a series circuit consists of only nodes each connected to only two elements, this means that the current in every point in a series circuit is the same. Kirchoff's voltage law: The sum of the signed voltage drops in a series circuit is zero. This means, that if you segregate the sources from the loads, the total voltage across all the nodes is equal to the total voltage across all the sources. That may seem trite, but take the case where you have one battery in series with two resistors also in series. If you know the voltage across one resistor, then you know the voltage across the other resistor - it is the battery voltage minus the first resistor's voltage. Ohm's law: Voltage is current times resistance. This actually applies everywhere; series circuits, parallel circuits, DC circuits, AC circuits, etc.
What is the max voltage for 3 50ohmn resisters?
It depends on the power rating of the resistors. The total power of a series or parallel combination of resistors is the sum of the power rating of each. Here is one possible answer that assumes the resistors are rated at 1/4 watt and are connected in series. Power = Current ^ 2 x Resistance. The ^ means squared. Current = square root (Power / Resistance) Current = square root (0.25 / 50) = 0.0707 amps The total power of the three resistors in series is 0.25 x 3 = 0.75 watts Current = square root (0.75 / 150) = 0.0707 amps <-- Notice you get the same current as before. This must be true because the current flowing through a series circuit is the same in each component. Since Voltage = Current x Resistance Voltage = 0.0707 x 50 = 3.54 volts across one of the resistors OR Voltage = 0.0707 x 150 = 10.61 volts across the series combination. Here's another way to calculate the answer. Voltage = square root of (Power x Resistance) Voltage = square root (0.25 x 50) Voltage = 3.54 volts This is the maximum voltage across one of the resistors. If the three resistors are connected in series, the total resistance would be 150 ohms and the maximum voltage across the series combination would be 3.54 x 3 = 10.61 volts. If the resistors are connected in parallel, the equivalent resistance is 16.67 ohms. Since the voltage across parallel resistors is the same, the maximum voltage for three 1/4 watt resistors would be square root (0.75 x 16.67) = 3.54 volts. This is the same answer calculated for the resistors in series. The maximum current through each resistor is V / R = 3.54 / 50 = 0.0707 amps or the same current as the series combination. However, in this case, the total current flowing through the parallel combination is 0.0707 x 3 = 0.2121 amps and Power = I ^ 2 x R = .2121 ^ 2 x 16.67 = 0.75W. This proves that the powers add no matter if the resistors are in series or parallel. If the resistors are rated at 1/2 watt the maximum voltage across one resistor is 5 V and the maximum across the series combination is 15 V. The maximum current is 0.1 A.
2ohm and 4ohm resistor are connected in series to a 12V battery what is the current through each resistor and use Ohm's law 2 show that the voltage drops across the individual resistors add up to 12V?
To answer this question, you need not only Ohm's law, but also Kirchoff's current and voltage Laws.Kirchoff's current law say the current at every point in a series circuit is the same. Lets call that current i.Ohm's law say voltage is resistance times current. That means the voltage across each resistor R1 and R2 is V1 = R1i and V2 = R2i.Kirchoff's voltage law says that the signed voltage drops around a series circuit add op to zero. This means that the voltages V1 and V2 must add up to be be equal to 12. This means that R1i + R2i = 12. This means that i = 12/(R1 + R2). But, hey, you say, this is just Ohm's law! This means that the current through the circuit is 2A. It also means that the resistance of two resistors in series is simply the sum of their resistance.Now that we know the current, we can calculate each resistor's voltage. V1 = R1i = (2)(2) = 4, and V2 = R2i = (4)(2) = 8. Crosscheck 4 + 8 = 12, so the two voltages do add up to twelve volts, as expected.In summary: The current through both resistors is 2A. The voltage across the 2 ohm resistor is 4V. The voltage across the 4 ohm resistor s 8V.
How you can prove the resistance in parqllel circuit?
The proof of the equivalent value of resistors in series lies in using Ohm's law along with Kirchoff's voltage and current laws.For a circuit of two resistors in parallel, across a voltage source, start with Kirchoff's voltage law. The signed sum of voltages drops across each element in a series circuit is zero. Said another way, the voltages across two parallel elements is the same. That means the voltage across each resistor is the same, and it is the same as the voltage across the voltage source.Since you now know the voltage across each resistor, you can calculate its current using Ohm's law.Now look at Kirchoff's current law. The signed sum of the currents entering a node is zero. Said another way, the current at every point in a series circuit is the same. Said yet another way, the way we are going to need it, is that the sum of the currents entering a node is equal to the sum of the currents leaving a node.You know the current in each resistor. Add them up and you get the current leaving the voltage source which is also the total current through the resistors. Knowing that current, and the voltage across them, go back to Ohm's law and plug in voltage and current, giving you net resistance.
What is the behavior of current and voltage in a series cricuit?
Kirchhoff's Current Law: The sum of the signed currents entering a node is zero. A consequence of this is that, in a series circuit every node only has two connections, one entering and one leaving, thus, in a series circuit, the current is the same at every point. Kirchhoff's Voltage Law: The sum of the signed voltage rises going around a series circuit is zero. This means, for example, that if you have two voltage drops, such as two bulbs in series with a battery, the voltage drops across them will add up to the voltage across the battery.
How does current react in a d.c. series circuit?
"DC" simply means that the voltage, and the current, goes in a single direction.
Which of kichhoff's laws applies to series circuits?
Kirchoff's Current Law states that the signed sum of the currents entering a node is equal to zero. In a simple parallel circuit, say with one battery and two light bulbs, this means the current coming out of the battery will be exactly equal to the sum of the currents entering the two light bulbs. In a series circuit, it also means that the current at every point in the circuit is the same. A parallel circuit can be construed as a special case of a series circuit, when you start to combine elements. Kirchoff's Voltage Law states that the signed sum of the voltage drops around a series circuit is equal to zero. Since a parallel circuit can be construed as a special case of a series circuit, this means that voltage across parallel nodes is equal.
How do parallel circuits use current and voltage?
Kirchoff's voltage law: In a series circuit, the signed sum of the voltage drops around the circuit add up to zero. Since a parallel circuit (just the two components of the parallel circuit) also represents a series circuit, this means that the voltage across two elements in parallel must be the same.Kirchoff's current law: The signed sum of the currents entering a node is zero. In a series circuit, this means that the current at every point in that circuit is equal. In a parallel circuit, the currents entering that portion of the circuit divide, but the sum of those divided currents is equal to the current supplying them.
