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If the resistance of the wire is 20 ohms and the voltage between the two ends of the wire is 20 volts,

then the current through the wire is

I = E/R = (40/20) = 2 amperes.

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14y ago
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15y ago

Use Ohm's Law V = I * R.

I = V / R = 12 / 4 = 3 [A].

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13y ago

v=ir

v=20volts

i=???

r=470 0hms

i=20/470=0.42 amps

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12y ago

I = E/R = 12/6 = 2 Amperes

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Q: How much current will a circuit need if it has twenty four ohms of resistance and a twelve volt power supply?
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A 120-V power supply connected to a 10-ohm resistor will produce how many amps of current?

120 volts into 30 ohms will produce 4 amps.Ohm's Law: Current equals Voltage divided by Resistance.Caution: Don't try this. The power in this case would be 480 watts, and an ordinary resistor would be destroyed by the heat generated, and might explode and catch fire.


How much current flows through a circuit if the if the resistance is 4 ohms and there is a 12-volt battery?

When two 4-ohm resistors are connected in parallel across 12 volts:-- Each resistor has 12V across it, so the current is I = E/R = 12/4 = 3 amperes.-- The total current from the battery is 6 amperes.-- The effective resistance as seen by the battery is R = E/I = 12/6 = 2 ohms.-- If you didn't know the voltage of the battery, you would calculate theeffective resistance of the parallel resistors in advance like this:1/R = 1/4 + 1/4 = 1/2R = 2 ohms.When you connected them to a battery, the battery would agree.


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.


Twelve identical wires of resistance 6 ohm each are arranged to form the edges of cube the effective resistance bw the opp corners of cube is?

The effective resistance between opposite corners of a cube comprised of twelve 6 ohm resistors, one at each edge, is 5 ohms. There are several ways to solve this. One approach is to build a system of 12 equations in 12 unknowns, and solve them. Another approach is this... Consider that there are three resistors leaving the input node, and there are three resistors entering the output node. In between those three resistors, there are six resistors in a criss-cross matrix. (Draw it out, flattened, to see this.)Inspecting the six resistors in the center, you note that they are completely symmetrical. Since they are symmetrical, you can conclude that the voltage at the junction between the three input resistors and the six others is the same voltage. The same goes for the three output resistors. Said another way, the voltage across the three input resistors and the three output resistors is the same. Given two or more nodes in a circuit having the same voltage, you can draw a wire connecting them, i.e. a resistor of zero ohms. This does not change the characteristics of the circuit in any way, because zero voltage across any resistance is still zero amperes. Now that you have made these connections, look at the circuit. It has simplified to three parallel resistors, in series with six parallel resistors, in series with three parallel resistors. Three 6 ohm resistors in parallel is 2 ohms. Six 6 ohm resistors in parallel is 1 ohm. Three more 6 ohm resistors in parallel is 2 ohms. The total resistance is 2 + 1 + 2 ohms, or 5 ohms.


How do you calculate the equivalent resistance of a circuit of the shape of a cube and a having a resistor of i kilo ohm in every branch?

Given twelve 1 KOhm resistors, connected in the shape of a cube, in order to determine the net resistance between opposite corners, first draw the cube in two dimensions. (Try this at each step before continuing, so you can understand the lesson as it unfolds.)There are three resistors leaving the initial vertex, and three resistors entering the final vertex. In between those six resistors, are six more resistors, each pair connected together on one end, and to two other resistors on the other end.If every resistor has the same value, then (by symmetry), the voltage on the ends of the first three resistors must be the same. Similarly, the voltage on the ends of the last three resistors must be the same.If two points in a circuit have the same voltage, then (for purposes of analysis) you can consider them to be shorted together. That short does not change the results, as there is no current flowing through that short.With the bottom ends of the first three resistors shorted, and with the top ends of the last three resistors shorted, the circuit degrades into three resistors in parallel, in series with six more resistors in parallel, in series with three more resistors in parallel.Three 1 KOhm resistors in parallel have a net resistance of 333 ohms. Six have a net resistance of 167 ohms. Two 333 ohm resistors and one 167 ohm resistor in series have a net resistance of 833 ohms, or 5/6 of 1 KOhms.Note: This technique does not work if the resistors are not all the same value. In that case, you would need to solve 12 equations in 12 unknowns, looking at the partial currents in each branch.

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If they are in series, you need to add the resistance values.


A 120-V power supply connected to a 10-ohm resistor will produce how many amps of current?

120 volts into 30 ohms will produce 4 amps.Ohm's Law: Current equals Voltage divided by Resistance.Caution: Don't try this. The power in this case would be 480 watts, and an ordinary resistor would be destroyed by the heat generated, and might explode and catch fire.


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When two 4-ohm resistors are connected in parallel across 12 volts:-- Each resistor has 12V across it, so the current is I = E/R = 12/4 = 3 amperes.-- The total current from the battery is 6 amperes.-- The effective resistance as seen by the battery is R = E/I = 12/6 = 2 ohms.-- If you didn't know the voltage of the battery, you would calculate theeffective resistance of the parallel resistors in advance like this:1/R = 1/4 + 1/4 = 1/2R = 2 ohms.When you connected them to a battery, the battery would agree.


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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.


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