Four 1000 ohm resistors in series have an effective resistance of 4000 ohms. Across a 4 volt voltage source, they would have a current of 1 mA, with a power dissipation of 4mW.
You just stated that the voltage across the resistor is 15 volts, so that's your answer ! If the resistor is connected to a 15-V battery or to the output of a 15-V power supply, then a meter across the resistor is also across the power supply, and reads 15 volts. The current through the resistor is (V/R) = (15/2700) = 5.56 mA. The power dissipated by the resistor (and delivered by the battery) is (V2/R) = (225/2700) = 0.083 watt.
A resistor&capacitor are in series with a variable inductor.When the circuit is connected to 200v,50Hz supply,the maximum current obtained by varying the inductance is 0.314A.The voltage across capacitor,when the current in circuit is maximum is 800v.Choose the values of series circuit elements?
At zero rpm a dc motor draws a current that could be enough to blow the circuit breaker. In that case it is normal to place a resistor in series to limit the current to slightly below the circuit-breaker setting. The resistor is switched out after the motor runs up.
16.32 volts
The 12V battery connected to the 2.4 Ohm combination will supply 12/2.4 or 5A. The individual currents will be 12/3 or 4A for the 3 Ohm resistor, and 12/12 or 1A for the 12 Ohm resistor. The 2.4 Ohm parallel combination is obtained from a simple product-over-the-sum calculation.
The protecting resistor is put in series with the LED so that you have a voltage divider - the supply voltage is split across the LED ( max 0.6v) and the remainder across the protecting resistor. So if your supply is 6volts, 5.4v will be across the resistor,
You just stated that the voltage across the resistor is 15 volts, so that's your answer ! If the resistor is connected to a 15-V battery or to the output of a 15-V power supply, then a meter across the resistor is also across the power supply, and reads 15 volts. The current through the resistor is (V/R) = (15/2700) = 5.56 mA. The power dissipated by the resistor (and delivered by the battery) is (V2/R) = (225/2700) = 0.083 watt.
Assuming it's 90 v dc, get a 1.5 k-ohm resistor and an 82 k-ohm resistor. Put them in series across the supply, then there will be slightly over 1.5 v across the smaller resistor.
If a 9V supply is connected to a 100-ohm resistor, then the current is not 2 Amps,or even close to it.I = E/R = 9/100 = 0.09 amp.
By using a voltage divider, that is two resistors of the same value in series across the DC supply. Half of the supply voltage will be at the point where the two resistors is connected. But how much wattage of those resistors is also an issue.
Well that's very interesting. What about it?
The equation for such circuits is given by: V=R*I +(1/C)*Integration(idt)
A resistor&capacitor are in series with a variable inductor.When the circuit is connected to 200v,50Hz supply,the maximum current obtained by varying the inductance is 0.314A.The voltage across capacitor,when the current in circuit is maximum is 800v.Choose the values of series circuit elements?
If there is nothing else in the circuit, then the voltage drop across the resistor will be the full supply voltage of 5 volts. The size of the resistor does not matter in this case - it will always be 5 volts.
At zero rpm a dc motor draws a current that could be enough to blow the circuit breaker. In that case it is normal to place a resistor in series to limit the current to slightly below the circuit-breaker setting. The resistor is switched out after the motor runs up.
LED interface requires typical 220 Ohm resistor in series While in case of relay resistor comes across coil with its value depends on supply voltage
The supply voltage in a parallel circuit remains the same regardless of the number of additional resistors connected. The voltage across each resistor in a parallel circuit is the same as the supply voltage. Adding more resistors in parallel will increase the total current drawn from the supply.