For the circuit above, assume that:𝑉=9𝑐𝑜𝑠(3𝑡)𝑉𝑅=4Ω𝐿=1𝐻𝐶=1𝐹Calculate the current flowing through the circuit, expressing your answer in the form: 𝐴𝑐𝑜𝑠(𝜔𝑡+𝜙), where 𝜙 is in degrees.Please round both𝐴 and 𝑝ℎ𝑖 to 3 sig figs, as otherwise the algebra-checker gets confused and awards you no points.

The given circuit is a series RLC circuit. The current in a series RLC circuit is given by the formula:

I = V / Z

where V is the voltage, and Z is the impedance of the circuit. The impedance of a series RLC circuit is given by the formula:

Z = sqrt(R^2 + (XL - XC)^2)

where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. The inductive reactance XL is given by the formula:

XL = ωL

and the capacitive reactance XC is given by the formula:

XC = 1 / (ωC)

Given that V = 9cos(3t), we can see that the angular frequency ω is 3 rad/s. The resistance R is 4 Ω, the inductance L is 1 H, and the capacitance C is 1 F. Substituting these values into the formulas, we get:

XL = ωL = 3 * 1 = 3 Ω XC = 1 / (ωC) = 1 / (3 * 1) = 0.333 Ω

Substituting R, XL, and XC into the formula for Z, we get:

Z = sqrt(R^2 + (XL - XC)^2) = sqrt(4^2 + (3 - 0.333)^2) = 4.717 Ω

Substituting V and Z into the formula for I, we get:

I = V / Z = 9 / 4.717 = 1.907 A

The phase angle φ is given by the formula:

φ = atan((XL - XC) / R)

Substituting XL, XC, and R into this formula, we get:

φ = atan((3 - 0.333) / 4) = 33.690 degrees

Therefore, the current flowing through the circuit is:

I = 1.907 cos(3t + 33.690) A