No. Only less.
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.
In order to determine what size of resistor is required to operate an LED from a 9V battery, first start by knowing the current and voltage required for the LED. That information is available in the LED's specifications. For discussion purposes, lets assume a typical LED at 2.5V and 50mW. The translates to a forward current of 20mA. Build a simple series circuit containing a 9V battery, a resistor of an as yet unknown value, and the LED. By Kirchoff's current law, the current in the LED is the same as the current in the resistor, which is also the same as the current in the battery. This is 20ma. By Kirchoff's voltage law, the voltage across the LED plus the voltage across the resistor equals the voltage across the battery. This is 6.5V. (9 - 2.5) By Ohm's law, resistance is voltage divided by current, so the resistor is 6.5 / 0.02, or 325 Ohms. The nearest standard value to that is 330 Ohms. Cross check the power through the resistor. Power is voltage times current, or 6.5V times 0.02A, or 0.13W. A half watt resistor is more than adequate for this job.
A voltmeter can be connected in parallel with a resistor to show the voltage across the resistor.
Increase the voltage across the resistor by 41.4% .
R = E/I = (12)/(0.1) = 120 ohms(Make it a big one. It dissipates I2R = 0.01 x 120 = 1.2 watts.)
v of what? v across what? v measured from what 2 points? v across the coils? v across the resistor? v across the coils and resistor? v across the battery? v across the battery and coils? v across the battery and resistor? or are you asking what v stands for? v stands for voltage.
The resistor is 1/3 of an ohm. A 9 volt drop across the resistor would cause a draw of 27 amps through the resistor. The wattage you would need for that resistor is at least a 243 watts.
wire a resistor across a battery. that is about as simple as it gets. the resistor could be an incandescent light bulb.
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.
Normally through the resistor's internal construction. It flows through any part of the resistor that has low resistance- be it anywere. And then there's this. It might be that one should consider that current flows through a resistor and voltage is dropped across a resistor. Perhaps this is where the question began. The former is fairly straight forward. The latter can be vexing. Voltage is said to be dropped across a resistor when current is flowing through it. The voltage drop may be also considered as the voltage measureable across that resistor or the voltage "felt" by that resistor. It's as if that resistor was in a circuit by itself and hooked up to a battery of that equivalent voltage.
A good AA battery has about 1.5 volts across it.
Yes
Volt across a resistor = resistance x current through the resistor.
You have two resistors, each with resistance of 12Ω, and a 12-volt battery. 1). The resistors are in series across the battery. ..... A. voltage across each resistor ..... B. current through each resistor ..... C. power dissipated by each resistor ..... D. total power delivered by the battery 2). The resistors are in parallel across the battery. ..... A. voltage across each resistor ..... B. current through each resistor ..... C. power dissipated by each resistor ..... D. total power delivered by the battery ============================================ 1). ... A. 6 volts ... B. 0.5 Amp ... C. 3 watts ... D. 6 watts 2). ... A. 12 volts ... B. 1 Amp ... C. 12 watts ... D. 24 watts
1). 6V battery, 1-ohm resistor, 2-ohm resistor, all in series:Total resistance = 3 ohms.Current in the loop = 6/3 = 2 amperesPower dissipated by the 2-ohm resistor - I2R = 8 watts.2). 4V battery, 12-ohm resistor, 2-ohm resistor, all in parallelThe 12-ohm resistor is irrelevant.4 volts across the 2-ohm resistor.Power dissipated by the 2-ohm resistor = E2/R = 8 watts.
sure whatever?
No. If a voltage is applied across a resistor, a current flows through it.