Respuesta :
(a) On the coil: 20 V, on the resistor: 0 V
The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:
[tex]V = V_R + V_L[/tex]
The potential difference across the inductance is given by
[tex]V_L(t) = V e^{-\frac{t}{\tau}}[/tex] (1)
where
[tex]\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s[/tex] is the time constant of the circuit
At time t=0,
[tex]V_L(0) = V e^0 = V = 20 V[/tex]
So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:
[tex]V_R = V-V_L = 20 V-20 V=0[/tex]
(b) On the coil: 0 V, on the resistor: 20 V
Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for
[tex]t >> \tau[/tex]
Since [tex]\tau[/tex] is at the order of less than milliseconds.
Using eq.(1), we see that for [tex]t >> \tau[/tex], the exponential becomes zero, and therefore the potential difference across the coil is zero:
[tex]V_L = 0[/tex]
Therefore, the potential difference across the resistor will be
[tex]V_R = V-V_L = 20 V- 0 = 20 V[/tex]
(c) Yes
The two voltages will be equal when:
[tex]V_L = V_R [/tex] (2)
Reminding also that the sum of the two voltages must be equal to the voltage of the battery:
[tex]V=V_L +V_R[/tex]
And rewriting this equation,
[tex]V_R = V-V_L[/tex]
Substituting into (2) we find
[tex]V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V[/tex]
So, the two voltages will be equal when they are both equal to 10 V.
(d) at [tex]t=5.77\cdot 10^{-4}s[/tex]
We said that the two voltages will be equal when
[tex]V_L=\frac{V}{2}[/tex]
Using eq.(1), and this last equation, this means
[tex]V e^{-\frac{t}{\tau}} = \frac{V}{2}[/tex]
And solving the equation for t, we find the time t at which the two voltages are equal:
[tex]e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s[/tex]
(e-a) -19.2 V on the coil, 19.2 V on the resistor
Here we have that the current in the circuit is
[tex]I_0 = 3.20 A[/tex]
The problem says this current is stable: this means that we are in a situation in which [tex]t>>\tau[/tex], so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law
[tex]V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V[/tex]
Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.
The voltage on the coil is given by
[tex]V_L(t) = -V e^{-\frac{t}{\tau}}[/tex] (1)
which means that it is maximum at the moment when the battery is disconnected, when t=0:
[tex]V_L(0)=.V[/tex]
And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.
(e-b) 0 V on both
After several seconds,
[tex]t>>\tau[/tex]
If we use this approximation into the formula
[tex]V_L(t) = -V e^{-\frac{t}{\tau}}[/tex] (1)
We find that
[tex]V_L = 0[/tex]
And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.
