Answer:
voltage divider, R₂ = 1000 R₁
measuring the output in the resistance R₁
Explanation:
Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V
in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.
To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.
If we use two resistors whose relationship is
R₂ / R₁ = 10³
R₂ = 1000 R₁
When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator
Answer:
f = 2 Hz
Explanation:
The frequency of a wave is defined as the no. of waves passing per unit of time. Therefore, the frequency of a wave can be calculated by the following formula:
where,
f = frequency of the wave = ?
t = time passed = 1 s
n = no. of waves passing in time t = 2
Therefore,
<u>f = 2 Hz</u>
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Answer:A i think or D but its not c or b
Explanation:
