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A 6μF capacitor that is initially uncharged is connected in series with a resistor and an emf source of 40 V with negligible internal resistance. The time constant is 30μs. At the instant when the resistor is dissipating electrical energy at a rate of 20 W, how much energy has been stored in the capacitor? Problem 3. (Full Score: 10 Points) A long, straight, cylindrical wire of radius R carries a current uniformly distributed over its cross section. At what locations is the magnetic field produced by this current equal to one third of its largest value? Consider points inside and outside the wire.

User Jjathman
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Answer:

The energy stored in a capacitor is given by:

U = (1/2) * C * V^2

where C is the capacitance and V is the voltage across the capacitor. At the instant when the resistor is dissipating electrical energy at a rate of 20 W, the power delivered to the capacitor must be equal to the power dissipated by the resistor, since the circuit is in steady state. Therefore:

P = V^2 / R = C * dV/dt * V^2

where R is the resistance, and dV/dt is the time derivative of the voltage across the capacitor. Solving for dV/dt, we get:

dV/dt = P / (C * V^2) = 20 W / (6 μF * V^2)

Since the capacitor is initially uncharged, we have V = emf = 40 V. Integrating both sides with respect to time, we get:

ln(V) - ln(V0) = -t / RC

where V0 is the initial voltage across the capacitor, and RC is the time constant. Solving for V, we get:

V = V0 * exp(-t / RC) = 40 V * exp(-t / 30 μs)

At the instant when the resistor is dissipating electrical energy at a rate of 20 W, the energy stored in the capacitor is:

U = (1/2) * C * V^2 = (1/2) * 6 μF * (40 V * exp(-t / 30 μs))^2

Substituting t = 0, we get:

U = (1/2) * 6 μF * (40 V)^2 = 4.8 mJ

So the energy stored in the capacitor at the instant when the resistor is dissipating electrical energy at a rate of 20 W is 4.8 mJ.

Regarding problem 3, the magnetic field produced by a long, straight, cylindrical wire of radius R carrying a current uniformly distributed over its cross section is given by:

B = μ0 * I / (2πr)

where μ0 is the permeability of free space, I is the current, and r is the distance from the wire. To find the locations where the magnetic field is equal to one third of its largest value, we need to solve the equation:

B = (1/3) * Bmax

where Bmax is the maximum

User Vuk
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