Question

Suppose that at any given time t (in seconds) the current j (in amperes) in an alternating current circuit is i=5 cos t + 5 sin t. What is the peak current for this circuit (largest magnitude)? The peak current for this circuit is amperes. (Type an exact answer, using radicals as needed.)

Answer 1

The peak current for the circuit is 5√2 amperes.

Step-by-step explanation:

The given equation for the current in the circuit is i = 5 cos t + 5 sin t. The peak current, or the largest magnitude of the current, can be found by finding the square root of the sum of the squares of the coefficients of the sine and cosine terms. In this case, the peak current is √(5^2 + 5^2) = √(50) = 5√2 amperes.

Answer 2

The peak current of the circuit, given by i(t) = 5 cos t + 5 sin t, is calculated using the trigonometric identity to combine sine and cosine terms. The peak current is found to be the square root of 50 Amperes.

Step-by-step explanation:

The student is asked to determine the peak current for an alternating current circuit whose current is given by the equation i(t) = 5 cos t + 5 sin t. To find the peak current, we need to express this equation in the form of i(t) = I0 sin(ωt + φ), where I0 is the peak current. We can do this by using trigonometric identities to combine the sine and cosine terms into a single sine term with an amplitude that represents the peak current.

First, we recognize that both sine and cosine terms have the same coefficient (5 in this case), so we can use the trigonometric identity which allows us to write A cos t + B sin t = R sin(t + φ), where R = √(A2 + B2) and φ is the phase angle such that tan(φ) = B/A. For our equation, A = B = 5, hence R = √(52 + 52) = √50. The peak current, I0, is therefore √50 Amperes, which is the largest magnitude of current the circuit will experience.