The Circuit contains a resistor R1, Resistor with R2 and an inductance L in series with a battery of emf ε0=V0 . R2 having a switch in parallel, the switch S is initially closed. At t = 0, the switch S is opened, so that an additional very large resistance R2 (with R2>>R1 ) is now in series with the other elements.
(a) If the switch has been closed for a long time before t = 0, what is the steady current I0 in the circuit? Express your answer in terms of, if appropriate, V0, R1, R2 and L
(b) While this current I0 is flowing, at time t = 0, the switch S is opened. Write the differential equation for I(t) that describes the behavior of the circuit at times t>0. Solve this equation (by integration) for I(t) under the approximation that V0=0 . (Assume that the battery emf is negligible compared to the total emf around the circuit for times just after the switch is opened.) Express your answer in terms of the initial current I0 , and R1 , R2 , t and L (for exponential function use e^(x) for ex).
(c) Using your results from part b), find the value of the total emf around the circuit (which from Faraday's law is −LdI/dt ) just after the switch is opened. Express your answer in terms of, if required, V0, R1, R2 and L
d) What is the magnitude of the potential drop across the resistor R2 at times t > 0, just after the switch is opened? Express your answers in units of V0 assuming R2=100R1
Did somebody figure out b) and c) please?
plz help.....what are the answers for a) n d) part ???
I got b) give d)
D= 100
anyways i need b and c :/ i went around in circles
To solve this circuit problem, we will follow these step-by-step guidelines:
(a) Steady current I₀ when the switch is closed for a long time:
Firstly, when the switch is closed for a long time, the inductor behaves like a wire, as there is no change in current. Therefore, the inductor can be considered a short circuit. In this case, the circuit consists of two resistors R₁ and R₂ in series with the battery, which leads to a simple series circuit.
Using Ohm's Law, we can find the current I₀. The total resistance in the circuit (R_total) is equal to R₁ + R₂ (since R₂ is closed). The steady current I₀ can be calculated using the formula: I₀ = V₀ / R_total, where V₀ is the emf of the battery.
(b) Differential equation for I(t) when the switch S is opened at t = 0:
When the switch S is opened, a large resistance R₂ (with R₂ >> R₁) is added in series with the other elements. We can write the differential equation for I(t) using Kirchhoff's voltage law.
The equation is: L(dI/dt) + R_total * I = 0. Here, L is the inductance, R_total is the total resistance in the circuit, and I is the current at time t.
To solve this differential equation, we'll rearrange it to isolate (dI/dt):
(dI/dt) = - (R_total / L) * I
Now, we can solve this separable first-order linear differential equation using integration.
∫(dI/I) = - (R_total / L) ∫dt
Integrating both sides, we get:
ln(I) = - (R_total / L) * t + C
Where C is the constant of integration. To determine C, we can use the initial condition I(t=0) = I₀.
ln(I₀) = C
Therefore, the final equation for I(t) is:
I(t) = I₀ * e^(- (R_total / L) * t)
(c) Total emf around the circuit just after the switch is opened:
Using Faraday's law (-L(dI/dt)), we can determine the total emf around the circuit. From part (b), we know that (dI/dt) = - (R_total / L) * I. Substituting this into Faraday's law, we have:
Total emf = -L * (dI/dt) = L * (R_total / L) * I
Total emf = R_total * I
Since we are assuming a negligible battery emf, the total emf will be due to the resistors, which is equal to R_total * I.
(d) Potential drop across resistor R₂ at t > 0 just after the switch is opened:
The potential drop across a resistor can be calculated using Ohm's Law. In this case, the current flowing through R₂ is I(t) as given in part (b).
The magnitude of the potential drop across R₂ is given by:
V_R2 = R₂ * I(t)
Given that R₂ = 100 * R₁, we have:
V_R2 = 100 * R₁ * I(t) = 100 * R₁ * I₀ * e^(- (R_total / L) * t)
Therefore, the potential drop across resistor R₂ at t > 0 just after the switch is opened is 100 * R₁ * I₀ * e^(- (R_total / L) * t) and it is expressed in terms of V₀, R₁, R₂, and L.