Finding voltages across capacitors under DC conditions

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SUMMARY

The discussion centers on calculating voltages across capacitors in a DC circuit. The circuit features three resistors in series and two capacitors, V1 and V2, which are treated as open circuits under DC conditions. The voltage V1 is determined to be 30V by calculating the current through a 30-ohm resistor, while V2 is derived from the equation V2 = 60 - 20*I1, where I1 is the current through a 20-ohm resistor, resulting in V2 being 40V. The analysis clarifies the relationship between voltage drops across resistors and the potentials at various points in the circuit.

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  • Basic concepts of capacitors in DC circuits
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ravensfan
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Homework Statement


Find the voltage across the capacitors in the circuit given under DC conditions.

Homework Equations


V=IR

The Attempt at a Solution


The solution is already given (the picture attached was obtained from the solution manual. I have a difficult time understanding how they arrived to that answer. V1 and V2 are really capacitors, but since this is in DC conditions, they are treated as open circuits.

I think I understand why V1 is 30v - since this circuit is essentially 3 resistors in series on the left side, all I'd need to do is find the current, then use that current across the 30 ohm resistor to find voltage, and since that resistor is in parallel with the capacitor, the voltage would be the same.

The right side is what I do not understand. Why would the voltage V2 be equal to 60-20*I1?

Thanks.
 

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ravensfan said:


The right side is what I do not understand. Why would the voltage V2 be equal to 60-20*I1?



Current I1=1A flows through the 20 ohm resistor, causing a potential drop of 20 V across it.
The potential at the positive terminal of the battery is 60 V with respect to the 0 point. Across a resistor, the current flows from the positive end towards the negative one, so the potential Ua is 20V lower than U+: Ua=60-20=40V. As no current flows through the 50 ohm resistor, the potential V2 is the same as Ua.
 

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ehild said:
Current I1=1A flows through the 20 ohm resistor, causing a potential drop of 20 V across it.
The potential at the positive terminal of the battery is 60 V with respect to the 0 point. Across a resistor, the current flows from the positive end towards the negative one, so the potential Ua is 20V lower than U+: Ua=60-20=40V. As no current flows through the 50 ohm resistor, the potential V2 is the same as Ua.

I understand what you are saying. Thank you very much for the help!
 

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