Question

You connect a battery, resistor, and capacitor as in (Figure 1), where R=17.0Ω and C=5.00×10 −6

F. The switch S is closed at t=0. When the current in the circuit has magnitude 3.00 A, the charge on the capacitor is 40.0×10 −6
C. What is the emf of the battery? Express your answer with the appropriate units. is Incorrect; Try Again; 5 attempts remaining Part B At what time t after the switch is closed is the charge on the capacitor equal to 40.0×10 −6
C ? Express your answer with the appropriate units. When the current has magnitude 3.00 A, at what rate is energy being stored in the capacitor? Express your answer with the appropriate units. Part D When the current has magnitude 3.00 A, at what rate is energy being supplied by the battery? Express your answer with the appropriate units.

Answer 1

The parts can be calculated as :

A) The emf of the battery is 510 V.

B) The time t when the charge on the capacitor is equal to 40.0×10-6 C is given by t = -(17.0 Ω) * (5.00×10-6 F) * ln(1 - (40.00×10-6 C)/(30.00 A * 5.00×10-6 F)).

C) The rate at which energy is being stored in the capacitor is (3.00 A) * (510 V) W.

D) The rate at which energy is being supplied by the battery is (3.00 A) * (510 V) W.

Given :

R = 17.0 Ω,

C = 5.00×10-6 F,

I = 30.00 A,

q = 40.00×10-6 C

A) To find the emf of the battery, we can use Ohm's Law. Since the current flow through the resistor is equal to the current flow through the capacitor, we have: I = V/R, where V is the emf. Solving for V gives us : V = I * R = (30.00 A) * (17.0 Ω) = 510 V.

B) To find the time t when the charge on the capacitor is equal to 40.0×10-6 C, we use the equation q = Q * (1 - e-t/RC), where Q is the maximum charge on the capacitor.

Plugging in the values, we have: 40.0×10-6 C = Q * (1 - e-t/RC).

Since the switch is closed at t=0, we can solve for t by rearranging the equation : e-t/RC = 1 - q/Q, where q/Q is the fraction of the charge on the capacitor. Taking the natural logarithm of both sides, we get : -t/RC = ln(1 - q/Q). Solving for t gives us: t = -RC * ln(1 - q/Q) = -(17.0 Ω) * (5.00×10-6 F) * ln(1 - (40.00×10-6 C)/(30.00 A * 5.00×10-6 F)).

C) The rate at which energy is being stored in the capacitor can be calculated using the equation P = IV, where P is power and I is the current. Since P = dU/dt, where U is the energy stored in the capacitor, we have : dU/dt = IV = (3.00 A) * (V) = (3.00 A) * (510 V). This gives us the rate at which energy is being stored in the capacitor in watts (W).

D) To find the rate at which energy is being supplied by the battery when the current has a magnitude of 3.00 A, we use the equation P = IV, where P is power and I is the current. Since P = IV, we have: P = (3.00 A) * (V), where V is the emf of the battery. Plugging in the value of V calculated in part A, we get the rate at which energy is being supplied by the battery in watts (W).