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The problem asks to find the steady state voltage across the capacitor. i attached the circuit (sorry for the poorly drawn and yes, that open part of the branch is supposed to be there) This is a multiple choice question where the answers are:

a) 0V

b) V_dc/2

c) V_dc/3

d) V_dc

e) 2*V_dc

In steady state, a capacitor behaves as an open.

If the capacitor was placed before the resistors, I think its voltage would be V_dc.

However, this capacitor is placed after the resistor.

In steady state, no current will flow through these 2 branches due to the open circuit, thus the voltage drop across the 2 resistors are 0.

This means that the ends of the capacitor are at the same voltage (V_dc). Thus the voltage difference is V_dc-V_dc=0.

So the answer is a. Am I right?

a) 0V

b) V_dc/2

c) V_dc/3

d) V_dc

e) 2*V_dc

In steady state, a capacitor behaves as an open.

If the capacitor was placed before the resistors, I think its voltage would be V_dc.

However, this capacitor is placed after the resistor.

In steady state, no current will flow through these 2 branches due to the open circuit, thus the voltage drop across the 2 resistors are 0.

This means that the ends of the capacitor are at the same voltage (V_dc). Thus the voltage difference is V_dc-V_dc=0.

So the answer is a. Am I right?