0
A 1k ohm and 3k ohm resistor are in series and the total power dissipated is 25 mW so what would the voltage drop across the 1k ohm resistor?
Wiki User
∙ 14y ago
1,000 ohms and 3,000 ohms in series = total effective resistance is (1,000 + 3,000) = 4,000 ohms.
Power dissipated = I2 R.
I = sqrt( P / R ) = sqrt (.025/4,000) = sqrt(6.25 x 10-6) = 0.0025 A
Voltage drop = I R
Across 1,000 ohms, V = 1,000 I = (1,000 x 0.0025)
= 2.5 volts.
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
How would you connect a resistor in order to reduce an LED's applied voltage?
You could use the voltage divider rule to reduce the voltage. Using two resistors in series, the input voltage will drop across each resistor by an amount that is proportionate to the values of the resistors. If the 1st resistor is 10K and the 2nd resistor is 100K, the voltage drop across the 10K will be 10 times LESS than that of the 100K resistor. The total voltage drop across both resistors will be equal to the supplied input voltage. Work out the ratio of voltage you need from the total input voltage and use 2 resistors will that give you the same ratio. Connect the LEDs in parallel with the resistor the gives you the voltage you want. Use a MM to measure the voltage across the resistor before wiring LEDs.
When resistors are connected in series in a circuit. what are the relationships between the voltage drops across the resistor and the currents through the resistors?
When resistors are connected in series in a circuit . the voltage drop across each resistor will be equal to its resistance, as V=IR, V is direct proportional to R. An A: The relationship is that the current will divide for each paths in a parallel circuit and the voltage drop across each will be the source voltage. In a series circuit the current will remain the same for each component but the voltage will divide to reflect each different component value. And the sum of all of the voltage drops will add to the voltage source
How do you find unknown value of a resistor when total resistor values and voltage are known in a series circuit?
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.
How much forward diode voltage is there with the ideal diode approximation?
Consider ideal diode to be connected in series with resistor of 6kSilicon diode forward bias voltage = 0.7 voltsCurrent across 6k resistor = (5-0.7)/6000 amperesVoltage across {resistor + diode}=4.3 + 0.7=5vIf silicon internal resistance is 6k then voltage across diode=5vIf external resistance is 6k and diode resistance is negligible then voltage across diode=0.7v
What happens to the voltage values of other resistors when a short occurs in one of them?
If a short occurs in a resistor in series with other resistors, the voltage drops across the other resistors will increase. If a short occurs in a resistor in parallel with other resistors, the voltage drops across the other resistors will decrease, to zero.
In a series circuit why does the largest voltage drop occur across the largest or smallest resistor?
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.
What is a protective resistor?
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,
When a circuit consists of a capacitor and a resistor in series with a voltage source and the voltage across the resistor becomes zero then the capacitor is?
fully charged.
Did the the total voltage across the resistors does not depend on the resistor value?
It depends on where and how the resistor is placed in a circuit. A string of series resistors will split the voltage across all them depending on their values. All of the resistors in parallel will have the same voltage across all of them no matter what their resistance is.
Is the heat loss and current of a resistor affected by being in a parallel circuit or can you just calculate it the same as in series?
The heat generated by any particular resistor depends (at least electrically) solely on the power it dissipates. Power dissipation in a resistor is equal to current squared times resistance, and the current through the resistor is equal to the voltage across it divided by the resistance. If we take a 10 ohm resistor ('your resistor') and put it in a series circuit such that there is 10 volts across your resistor, the current through it will be 1 ampere (10/10=1). the power dissipated will be 10 watts (1^2 * 10=10). If we put your resistor in a parallel circuit that also puts 10 volts across it, then the current and power will be the same. Your resistor does not know or care where the voltage came from. From this point of view, once you get down to the voltage across the resistor, it does not matter what type of circuit it is in. On the other hand, for any given power supply voltage, then the type of circuit and the value of external components certainly does affect the terminal voltage and thus the current through as well as the power dissipated by the resistor. In a parallel circuit, the voltage across your resistor remains basically the same no matter what resistance you put in parallel with it (unless you overload the power supply or the power supply has high internal resistance). In this case, the voltage across the resistor is the same as the power supply, current is I=E/R, R being that resistor only, and power is P=I^2 * R. In a series circuit the current through the resistors is I=E/R, R being the total resistance (including the other resistor(s)). The power dissipation in your resistor will then be P=I^2 * R, I being the series current we just calculated, and R being your resistor only. Since the other resistors affect the current, and since the current is the same no matter where you measure in a series circuit, then the voltage across your resistor and thus the power dissipation will be affected. The voltage across your resistor will be E=I*R, I being the series current we just calculated, and R being your resistor only. So, while the calculation for power dissipated in a particular resistor does not change relative to what type of circuit it is in, the calculation to arrive at the voltage across the resistor and/or the current through it (which you will then need to calculate power) does. Keep in mind there are other mechanical parameters that influence the actual case temperature of the resistor. Physical size of the case, composition, and airflow velocity, if any, will alter the case-to-ambient thermal conductivity. Ambient temperature will also be a factor in the final temperature.
If the resistance in the circuit is increased what will happen to the current and voltage?
* resistance increases voltage. Adding more resistance to a circuit will alter the circuit pathway(s) and that change will force a change in voltage, current or both. Adding resistance will affect circuit voltage and current differently depending on whether that resistance is added in series or parallel. (In the question asked, it was not specified.) For a series circuit with one or more resistors, adding resistance in series will reduce total current and will reduce the voltage drop across each existing resistor. (Less current through a resistor means less voltage drop across it.) Total voltage in the circuit will remain the same. (The rule being that the total applied voltage is said to be dropped or felt across the circuit as a whole.) And the sum of the voltage drops in a series circuit is equal to the applied voltage, of course. If resistance is added in parallel to a circuit with one existing circuit resistor, total current in the circuit will increase, and the voltage across the added resistor will be the same as it for the one existing resistor and will be equal to the applied voltage. (The rule being that if only one resistor is in a circuit, hooking another resistor in parallel will have no effect on the voltage drop across or current flow through that single original resistor.) Hooking another resistor across one resistor in a series circuit that has two or more existing resistors will result in an increase in total current in the circuit, an increase in the voltage drop across the other resistors in the circuit, and a decrease in the voltage drop across the resistor across which the newly added resistor has been connected. The newly added resistor will, of course, have the same voltage drop as the resistor across which it is connected.
How does voltage and resistance behave in a series and a parallel circuit?
The current through each resistor is equal to the voltage across it divided by its resistance for series and parallel circuits.
How would you connect a resistor in order to reduce an LED's applied voltage?
You could use the voltage divider rule to reduce the voltage. Using two resistors in series, the input voltage will drop across each resistor by an amount that is proportionate to the values of the resistors. If the 1st resistor is 10K and the 2nd resistor is 100K, the voltage drop across the 10K will be 10 times LESS than that of the 100K resistor. The total voltage drop across both resistors will be equal to the supplied input voltage. Work out the ratio of voltage you need from the total input voltage and use 2 resistors will that give you the same ratio. Connect the LEDs in parallel with the resistor the gives you the voltage you want. Use a MM to measure the voltage across the resistor before wiring LEDs.
When resistors are connected in series in a circuit. what are the relationships between the voltage drops across the resistor and the currents through the resistors?
When resistors are connected in series in a circuit . the voltage drop across each resistor will be equal to its resistance, as V=IR, V is direct proportional to R. An A: The relationship is that the current will divide for each paths in a parallel circuit and the voltage drop across each will be the source voltage. In a series circuit the current will remain the same for each component but the voltage will divide to reflect each different component value. And the sum of all of the voltage drops will add to the voltage source
What is series connection?
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
How do you find unknown value of a resistor when total resistor values and voltage are known in a series circuit?
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.
How is a parallel circuit the same as a series circuit?
In both cases, the power dissipated is measured by multiplying the voltage across the circuit by the current through the circuit.
