if a voltage of 20 cos 5t is applied to a series circuit consisting of 10 ohm resistor and 2 henry inductor determine the current at any time t>0 hint (di/dt)+(r/l)i=(e(t)/l)

To determine the current at any time t>0 in the given series circuit, we can use the differential equation (di/dt) + (R/L)i = (E(t)/L), where R is the resistance, L is the inductance, i is the current, and E(t) is the applied voltage.

Given: Voltage, E(t) = 20 cos(5t) Resistance, R = 10 ohms Inductance, L = 2 henries

We can substitute these values into the differential equation and solve for the current.

(dI/dt) + (10/2)I = (20 cos(5t))/2

Simplifying the equation, we have:

(dI/dt) + 5I = 10 cos(5t)

This is a first-order linear ordinary differential equation. To solve it, we can use an integrating factor.

The integrating factor, μ(t), is given by:

μ(t) = e^(∫5 dt) = e^(5t)

Multiplying both sides of the equation by μ(t), we get:

e^(5t)(dI/dt) + 5e^(5t)I = 10e^(5t) cos(5t)

Now, we can rewrite the left side of the equation using the product rule:

(d/dt)(e^(5t)I) = 10e^(5t) cos(5t)

Integrating both sides with respect to t, we have:

∫(d/dt)(e^(5t)I) dt = ∫10e^(5t) cos(5t) dt

e^(5t)I = ∫10e^(5t) cos(5t) dt

To evaluate the integral on the right side, we can use integration by parts. Let's assume u = cos(5t) and dv = 10e^(5t) dt.

Differentiating u, we get du = -5 sin(5t) dt Integrating dv, we get v = 2e^(5t)

Using the integration by parts formula, we have:

∫u dv = uv - ∫v du

Substituting the values, we get:

∫10e^(5t) cos(5t) dt = 2e^(5t) cos(5t) - ∫2e^(5t)(-5 sin(5t)) dt

Simplifying further, we have:

∫10e^(5t) cos(5t) dt = 2e^(5t) cos(5t) + 10∫e^(5t) sin(5t) dt

Now, we can use integration by parts again. Let's assume u = sin(5t) and dv = e^(5t) dt.

Differentiating u, we get du = 5 cos(5t) dt Integrating dv, we get v = (1/5)e^(5t)

Using the integration by parts formula, we have:

∫e^(5t) sin(5t) dt = -(1/5)e^(5t) cos(5t) + ∫(1/5)e^(5t)(5 cos(5t)) dt

Simplifying further, we have:

∫e^(5t) sin(5t) dt = -(1/5)e^(5t) cos(5t) + ∫e^(5t) cos(5t) dt

Now, we can substitute this result back into the previous equation:

∫10e^(5t) cos(5t) dt = 2e^(5t) cos(5t) + 10(-(1/5)e^(5t) cos(5t) + ∫e^(5t) cos(5t) dt)

Simplifying further, we have:

∫10e^(5t) cos(5t) dt = 2e^(5t) cos(5t) - 2e^(5t) cos(5t) + 10∫e^(5t) cos(5t) dt

∫10e^(5t) cos(5t) dt = 10∫e^(5t) cos(5t) dt

Now, we can substitute this result back into the original equation:

e^(5t)I = 2e^(5t) cos(5t) + 10∫e^(5t) cos(5t) dt

Dividing both sides by e^(5t), we get:

I = 2 cos(5t) + 10∫cos(5t) dt

Integrating cos(5t) with respect to t, we have:

I = 2 cos(5t) + 10(1/5) sin(5t) + C

Simplifying further, we have:

I = 2 cos(5t) + 2 sin(5t) + C

Therefore, the current at any time t>0 in the given series circuit is given by the equation I = 2 cos(5t) + 2 sin(5t) + C, where C is the constant of integration.