Question

Build this circuit in tinkercad with R₁=100Ω,R₂=1kΩ and C=1μF, using the function generator for V₀,

with zero DC olfset, 1 volt amplitude and frequency, f, at the values below,
and use the oscilloscope to measure V₁ in peak-to-peak volts.
What is V₁ when f=40 Hz ?

Answer 1

In the given circuit with R₁ = 100Ω, R₂ = 1kΩ and C = 1μF, the voltage V₁ is approximately 0.978 V when the frequency is 40 Hz.

Step-by-step explanation:

In the given circuit with R₁ = 100Ω, R₂ = 1kΩ and C = 1μF, and when the frequency is 40 Hz, we need to find the voltage V₁. The circuit consists of a series connection of two resistors with a capacitor. The voltage V₁ is the voltage across the second resistor, R₂. In a series circuit, the total impedance (Z) is given by the formula:

Z = sqrt((R₁+R₂)² + (1/(2πfC))²)

Using the given values, we can calculate the impedance and find the current (I) using Ohm's Law, I = V₀ / Z, where V₀ is the function generator voltage. Finally, we can find V₁ by multiplying the current with R₂ since V₁ = I * R₂.

Let's calculate the voltage V₁:

Z = sqrt((100+1000)² + (1/(2π*40*10^-6))²) ≈ 1022.36Ω

I = 1 V / 1022.36Ω ≈ 0.000978 A

V₁ = 0.000978 A * 1000Ω ≈ 0.978 V