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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.
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A: If you know the total resistance and total voltage then you know total current flow for the circuit, this current will be same for every resistor in series however the voltage drop will change for each resistor . So measuring the voltage drop across the resistor in question and divide by the total current will give you the resistor value.
if R4 is the only resistor (the load), then the drop would be the same as the energy source
There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
A: Simply by adding a series resistor from the battery charger. WHAT VALUE? find the current required and use it to IR drop the voltage
It is a meter with a high wattage resistor connected parallel with the meter that drains power from the battery while it is measuring the voltage it is used to test the condition of the battery if the condition is good there will not be a large voltage drop while testing the battery
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.
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.
What is the voltage drop running through resistor one
The correct question is what is the voltage drop across a resistor or the current flowing through the resistor using Ohm's Law where Voltage = Current x Resistance
It doesn't. In a series circuit, the largest voltage drop occurs across the largest resistor; the smallest voltage drop occurs across the smallest resistor.
The resistor with the most resistance.
What is the amount of current flowing through the resistor? Voltage drop is dependent on the current. Ohm x Amps = Voltage drop
No, it would overcharge and ruin the battery. #2 You can do it by using a series resistor to limit the current. The resistance needed is equal to the voltage drop divided by the charging current. The voltage drop is 4, and the charging current is one tenth the battery capacity, for example for a 2000 mAh battery the charging current is 200 mA. So the resistance needed is 4/0.2 or 20 ohms. The power dissipated in the resistor is 4 x 0.2, or 0.8 watts, so you need a 20 ohm 1-watt resistor. The battery should be charged for at least 10 hours if fully discharged, or until it becomes slightly warm to the touch. Do NOT let it overheat.
Any part of a circuit that has a voltage drop across it is a resistor.
A resistor drops both voltage and current, however the term "drop" is generally used to indicate a voltage or current drop across the device, so it is more correctly stated that a resistor drops voltage, by allowing the current in the circuit to decrease.