How do I calculate the voltage across each capacitor after the switch is closed?

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To calculate the voltage across each capacitor after the switch is closed, it is essential to understand that the circuit reaches a steady state where the current through the capacitors is zero. The voltage across the 5 µF capacitor is 20 V, equal to the voltage across the 40-ohm resistor, due to the principles of voltage division in the circuit. For the 10 µF capacitor, the voltage can be determined by analyzing the resistive voltage divider in the circuit. After a long time, the capacitors behave as open circuits, allowing for straightforward calculations based on the resistances present. Understanding these concepts is crucial for accurately determining the voltages across capacitors in a DC circuit.
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Homework Statement



Personal question: What is the voltage across each capacitor after the switch is closed?
(It was previously calculated that the current after the switch is closed is .5 A)

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Homework Equations



Q = CV
Q = Qo*e^(-t/RC)
V = IR

The Attempt at a Solution



The answer is 20 V for the 5uF capacitor, but I do not understand why the voltage for the 5 uF capacitor is the same as the voltage across the 40 ohm resistor.

I have no idea how to calculate the voltage for the 10 uF capacitor.
 
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Presumably you're looking for the voltage across the capacitors a "long time" after the switch is closed, after the transients have died away and everything is static?

Consider that at that time the currents through the capacitors will all be zero (we've reached steady state for a DC circuit) , so as far as the resistance network around the capacitors is concerned, the capacitors no longer exist. See if you can identify a resistive voltage divider that's setting the final voltages at key points in the circuit.
 
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