To solve this problem, let's break down the circuit behavior at two specific points in time:

Part (a): Immediately after the switch closes

When the switch closes, the capacitors are initially uncharged, which means they act like short circuits (their voltage is zero initially, so they do not impede the current). Therefore, the current flows through all paths that include the capacitors as if they were regular wires.

1. The 100 V battery will drive current through the resistive components only.
2. Calculate the total effective resistance immediately after the switch is closed.
   • Combine resistances as needed, treating capacitors as short circuits.

Part (b): After a very long time (capacitors fully charged)

When capacitors are fully charged, they act like open circuits because no steady-state DC current can flow through a fully charged capacitor.

1. Remove the branches with capacitors from the circuit since they block DC current.
2. Calculate the effective resistance of the remaining resistors in series or parallel as appropriate.

I will perform the calculations based on these steps and assumptions.from sympy import symbols, Eq, solve

Given circuit values

V_battery = 100 # Voltage of the battery in volts R1 = 25 # Resistance in ohms R2 = 75 # Resistance in ohms R3 = 50 # Resistance in ohms R4 = 15 # Resistance in ohms R5 = 25 # Resistance in ohms (two in parallel)

(a) Immediately after the switch closes

Capacitors act as short circuits initially

Calculate the equivalent resistance for the resistors in the circuit with capacitors shorted

First, R5 and another R5 in parallel

R_parallel_5 = 1 / (1 / R5 + 1 / R5)

Then, add the parallel combination with R3 and R4 in series

R_total_initial = R1 + R2 + R_parallel_5 + R3 + R4

Using Ohm's law I = V / R

I1_initial = V_battery / R_total_initial

(b) After a very long time (capacitors fully charged, treated as open circuits)

In this case, we remove branches containing capacitors

Calculate the equivalent resistance with capacitors as open circuits

R3 and R5 (parallel combination) in series with R4

R_parallel_3_5 = 1 / (1 / R3 + 1 / R5) R_total_final = R1 + R_parallel_3_5 + R4

Using Ohm's law I = V / R

I2_final = V_battery / R_total_final

I1_initial, I2_final(0.5633802816901409, 1.7647058823529411)(0.5633802816901409, 1.7647058823529411)successThe readings on the ammeter are as follows:

1. Immediately after the switch closes (I1): $0.563 \, \text{A}$
2. After a very long time, when capacitors are fully charged (I2): $1.765 \, \text{A}$

Would you like further details on the calculations, or do you have any specific questions?

Related Questions:

1. How would the current change if the battery voltage were increased to 200 V?
2. What would happen to the current if the capacitance of each capacitor was doubled?
3. How would adding another resistor in series with $R_1$ affect the initial and final current?
4. What if the capacitors had an initial charge before the switch was closed?