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What does a voltage divider do and how does it work to divide the input voltage into smaller output voltages?
A voltage divider is an electrical circuit that divides the input voltage into smaller output voltages. It consists of two resistors connected in series. The output voltage is determined by the ratio of the two resistors. The formula for calculating the output voltage is Vout Vin (R2 / (R1 R2)), where Vin is the input voltage, R1 is the resistance of the first resistor, R2 is the resistance of the second resistor, and Vout is the output voltage.
How do you calculate output voltage?
The output voltage can be calculated using Ohm's Law, which states that V (voltage) = I (current) * R (resistance), where I is the current flowing through the circuit and R is the resistance of the circuit. Alternatively, you can use the voltage divider formula Vout = Vin * (R2 / (R1 + R2)), where Vin is the input voltage and R2 / (R1 + R2) is the voltage divider ratio of resistors R1 and R2.
What happen to DC level when Voltage-Division is adjusted to any value?
When the voltage division is adjusted in a circuit, the DC level at the output of the voltage divider changes according to the values of the resistors used and the input voltage. Essentially, the output voltage can be calculated using the formula ( V_{out} = V_{in} \times \frac{R2}{R1 + R2} ), where ( R1 ) and ( R2 ) are the resistances in the divider. As the resistor values or the input voltage change, the DC level at the output will also vary proportionally. This adjustment allows for precise control of the DC voltage level delivered to subsequent circuit components.
How does work the op amp?
An op-amp is a phase-inverting voltage amplifier (high input-impedance, low output-impedance) with a large voltage gain. Shunt feedback is connected through an impedance between the output and input terminals. In the simplest configuration there is a shunt feedback resistor - call that R2. If another resistor R1 is conncted in series with the input signal, the op-amp then produces a voltage gain of R2/R1. The new voltage gain is considerably less than the gain without any feedback, and the practical result of this is that the input terminal of the op-amp, which is also the junction of the two resistors, always has an extremely low signal voltage on it and it is termed a 'virtual earth'. The operation of the circuit can be considered by having the signal voltage inducing a signal current equal to Vin/R1 through R1, which then flows through R2 to produce a signal voltage of Vin R2/R1. If the feedback resistor R2 is replaced by a capacitor C, then the gain of the amplifer is the ratio of the capacitor's impedance and the input resistor, R, or Vout/Vin = 1/jwCR. This is know as an 'integrator' because the output voltage is the time integral of the input voltage: it has a phase-lag of 90 degrees and an amplitude that decreases at the rate of 6 dB per octave. If a constant current of i amps is applied to the input terminal, the output voltage rises on a ramp with a slope of i/CR volts per second; and if a variable current is applied, the output voltage is 1/(CR) integral i.dt. This is why an op-ap with a feedback capacitor and a series resistor i known as an analogue integrator.
What are two reasons for the absence of R1 in the output?
Refer to the exhibit. A network administrator successfully pings R1 from R3. Next, the administrator runs the show cdp neighbors command on R3. The output of this command is displayed.What are two reasons for the absence of R1 in the output? (Choose two.)~**The no cdp run command has been run at R1.**The no cdp enable command has been run at Fa0/1 interface of R3.R1 is powered off.
What will be the resulting voltage if a 4.5 voltage source is connected in parallel to a 5.0 volts source?
Theoretically this cannot happen because a voltage source isn't practically available. If you managed to find near perfect voltage sources of differing voltages and put them in parallel then you'd get some massive currents flowing. Why? You have two sources each wanting there own voltage to exist across itself, by putting them in parallel they will practically settle at a voltage somewhere between the two. This settling voltage will be determined by the internal resistances of each source. If R1=internal resistance of the 4.5V source and R2=internal resistance of the 5V source. Current flow=Voltage difference / resistance. So current flow=0.5/(R1+R2) Settling Voltage = 4.5 + (current *R1) = 4.5+0.5*(R1/(R1+R2)) So if R1=0.001 and R2=0.002 Ohms Current = .5/(.003) = 167 Amps Voltage = 4.5 + .5*(.001/.003) = 4.67V. This explains why you shouldn't put power batteries in parallel, there is a risk of massive currents to flow. Andrew
What happens to the current of R1 connected in series with R2 when it is also connected in parallel with R3?
The current, if connected to a voltage source that can supply the needed current to (R1+R2) R3, will be unchanged. If the source cannot supply the needed current, the terminal voltage will decrease, which will change the current flowing through R1 and R2.
What would the tracert output be if r1-isp failed?
Nothing reach after r2 central.
How do you calculate the ir1 and ir2 in a parallel circuit?
To calculate the currents ir1 and ir2 in a parallel circuit of two resistors r1 and r2, consider the voltage across them. By Kirchoff's voltage law, the voltage across each resistor will be the same, so simply use Ohm's law to divide voltage by each resistor to get each resistor's current. ir1 = v / r1 ir2 = v / r2 If you don't know voltage, but you do know total current, then determine the total resistance as RT = R1R2/(R1+R2), calculate voltage as ITRT and proceed from there.
What is the Voltage drop of R1 If V 10 Volts R1 500 Ohms and R2 1500 Ohms?
You need to mention whether the resistors are in series or parallel to get the right answer
What is the total current of the circuit if voltage is 12 R1 is 100 Ohm and R2 is 200 Ohm?
As you don't specify how R1 & R2 are connected, it is impossible to give an answer to your question.
What will happen to the current through R1 and R2 if R3 fails open?
If R3 fails open, it effectively removes it from the circuit, increasing the total resistance. As a result, the current through R1 and R2 will decrease because the total current flowing in the circuit will be reduced due to the higher resistance. Depending on the configuration of the resistors, the voltage across R1 and R2 may also change, further affecting the current through each resistor.
