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.)
In a direct current (DC) circuit, a capacitor will eventually charge up and act as an open circuit, meaning it will not allow current to flow after reaching full charge. As a result, the impedance of a resistor-capacitor (C-R) circuit under DC conditions is simply the resistance value. Therefore, the impedance of the given C-R circuit with a resistance of 20 ohms and a capacitance of 2 microfarads is 20 ohms.
Using Ohms Law: V = I x R, where V (Voltage), I (Current), and R (Resistance). re-arranging: V/R = I Therefore if you double both the Voltage and the Resistance, the current remains unchanged.Current = Voltage / Resistance. If both resistance and voltage double the current remains the same.
Your original question was in two parts:1.) How many ohms in an open circuit? Infinite ohms (the meter will show no measurement).2.) How many ohms in a short circuit? 0 ohms. There would be no measurable ohms as there would be no resistance in the altered circuit.
When a wire is cut in a circuit, a gap is made and the current can no longer circulate, known as an open circuit.When 2 parts of a circuit touch, that shouldn't, for example - a wire comes loose and comes into contact with another part of the circuit, its shortening the route of the current in the circuit. So its a short circuit. When this happens 99.9% of the time the result will be a spike in amp's, so tripping any circuit protection, MCB's, fuses.A good example of a common short circuit is faulty windings on a 3-phase electric motor. If the resin separating the windings becomes damaged it can cause 2 or even 3 of the motors phases to come into contact causing the motors overload protection to trip.
P=I^2*R. No. 8,000 watts.
by adding the the resistances in series the total resistance of the circuit increses and thus the crunt flowing in the circuit decrese. Ans 2 . the current in series circuit of constant resistance will always be the same . It will not effect the current .
To find the current in the circuit, you can use the formula: Power = Current^2 * Resistance. Given the values, you can rearrange the formula to solve for current: Current = sqrt(Power / Resistance). Plugging in the values, you get Current = sqrt(2 / 30) which simplifies to approximately 0.27 amperes.
If we apply Ohm's law, which is E = I x R and we have a voltage (E) of 110 volts and a current (I) of 10 amps, we can use the variation of the formula to solve. That variation is R = E / I and the resistance (R) is discovered by dividing the voltage by the current. R = E / I = 110 / 10 = 11 ohms
V=IR where V is voltage, I is current and R is resistance. You want to know what the current will be in a series circuit based on the resistance. You need to know the voltage as well as the resistance, gives you the equation as follows I=V/R So if you have 10 volts and a 1 ohm resistor, the current will be 10 amps. If you increase the resistor to 10 ohms, your current will then be 1 amp. In a parallel circuit, the resistance is equal to the sum of the inverse. For example. If I have two resistors of 2 ohms each in parallel, the equation would be 1/2 + 1/2 = 0.5 + 0.5 = 1 In that particular instance, your current would increase.
No, power is not directly proportional to resistance. The power dissipated in a circuit is given by P = I^2 * R, where I is the current flowing through the circuit and R is the resistance. This means that power is proportional to the square of the current but linearly proportional to resistance.
Well, first of all, if the resistance of the circuit is 10 ohms and you connect 10 volts to it,then the current is 1 Amp, not 2 . So either there's something else in your circuit thatyou're not telling us about, or else the circuit simply doesn't exist.-- If you connect some voltage to some resistance, then the resistance heats up anddissipates (voltage)2/resistancewatts of power, and the power supply has to supply it.-- If there is some current flowing through some resistance, then the resistance heats up anddissipates (current)2 x (resistance)watts of power, and the power supply has to supply it.-- If there's a circuit with some voltage connected to it and some current flowingthrough it, then the resistance of the circuit is (voltage)/(current) ohms, the partsin the circuit heat up and dissipate (voltage) x (current) watts of power, andthe power supply has to supply it.There's no such thing as "the power of a circuit". The power supply supplies thecircuit with some amount of power, the circuit either dissipates or radiates someamount of power, and the two amounts are equal.
There are two possible causes: 1. The circuit has no Voltage applied to it. 2. The resistance of the circuit is INFINITE.
I don't know what the parallel circuit has to do with it. You've onlygiven me a resistor and the current through it.When 0.03A of current passes through a 1,000Ω resistor, the resistordissipates energy at the rate of 0.9 watt.
Ohm's law is V = I·R. You know V and I, so you can calculate R using R = V/I.60 V / 2 A = 30 Ω
Series circuit: The total voltage is the sum of the voltage on each component. The total resistance is equal to the sum of the resistance on each component. The total current is equal in every component.
Without knowing permissible current draw by the divider or its maximum power dissipation the actual resistor values cannot be determined, but the ratios of the resistor values can be determined from the required voltage drops.The divider will be composed of 4 resistors starting at the 10VDC rail:2VDC drop, ratio = 2V/10V = 0.23VDC drop, ratio = 3V/10V = 0.33VDC drop, ratio = 3V/10V = 0.32VDC drop, ratio = 2V/10V = 0.2Therefore you will need 2 resistors (R1 & R4) that are 0.2 * the total resistance of the voltage divider and 2 resistors (R2 & R3) that are 0.3 * the total resistance of the voltage divider.But as stated at the beginning you can get no further without additional requirements being specified.
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