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Surely you must realize that it's impossible to answer that question without seeing
the schematic diagram of the circuit. Perhaps if you glanced at it yourself, you might
not need any help answering the question.
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∙ 11y ago
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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 would the tracert output be if r1-isp failed?
Nothing reach after r2 central.
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.
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 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 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
Derive the output equation for differential amplifier and instrumentation amplifier?
Vo=(R2/R1)(V2-V1)
Why there is algebric sum of the resistances in the series combination?
Ohms Law will be helpful in seeing how resistances add up. Let's assume you have a 10 ohm and a 20 ohm resistor in series and 30 Volts. across the series. Ohm's Law states that Voltage = Resistance x Current. If we describe the 10 ohm resistor as R1 and the other as R2 then the voltage drop across R1 is V1 and V2 is the drop across R2. This can be written V1 = R1 x I1 and V2 = R2 x I2. Since the total voltage must equal the sum of the voltage drops then Vtot = V1 + V2. Also Itot = I1 + I2. Substituting we get Vtot = (I1 x R1) + (I2 x R2) = (I1 + I2) x (R1 + R2). And Vtot = Itot x Rtot so Rtot = R1 + R2. In example 30 Volts = Itot x (10 + 20) or Itot = 1 amp.
Are there any cheat codes for madden 09 for PS2?
yes press R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2,R1,R2, then press select to complete the entire game
What is the main use of ic 7812?
Here is circuit 9V power supply regulator with battery 12V backup system. Use IC 7812 and 7805 control voltage output by R3 1K output 9V 1A max. This is a 9V power supply which will work even on power failure. It uses a rechargeable battery and regulators. A transformer with 15-0-15 AC volts output is required. In the first regulator U1 the output is lifted up by 1.4V and in the second regulator U2 by a resistor divider. In the second regulator the voltage across resistor R3 is 5V, so the current is 5V / 1K = 5mA this adds to the quiescent current of 5mA from the regulators ground terminal and flows into the resistors R1 and R2 in parallel which form 404 ohms, 10mA thru 404 ohms is 4V. So the output will be 5 + 4 = 9V. Note that the charge and discharge paths of the battery are separated with diodes.
