Finding voltages across capacitors under DC conditions

AI Thread Summary
In a DC circuit with capacitors, the capacitors are treated as open circuits, which simplifies the analysis. The voltage across capacitor V1 is determined to be 30V by calculating the current through a series of resistors. For capacitor V2, the voltage is derived from the battery's potential, taking into account the voltage drop across a 20-ohm resistor due to the current flowing through it. The potential at the point before the resistor is 60V, and after the drop, it is 40V, which is also the voltage across V2 since no current flows through the 50-ohm resistor. This understanding clarifies the relationship between the components in the circuit.
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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