Why is the voltage $V_c$ equal to the voltage source?

As I said in the title why is $$\V_c = V_{\text{in}} = 12\$$V? I tried applying Kirchhoff's voltage law (not sure whether I'm allowed), assuming that the current of the circuit moves clockwise I get : $$-V_{\text{in}} + R_1 i + V_c = 0$$ What I'm wondering is whether there's a current flowing since this circuit open. Where'd the current go? I assume there's a current due to the voltage source. I'm very likely wrong, any input would be greatly appreciated.

• Current doesn't flow in an open circuit. So what would be the voltage across the resistor? KVL is absolutely applicable, so given no current flow, you should be able to see why Vc is equal to Vin. Commented Sep 2 at 18:35
• A bit handwavy for an answer but you can replace the open circuit with a resistor of an arbitrarily large value and proceed from there.
– vir
Commented Sep 2 at 18:43

Your application of KVL is correct. now coming to your question, the circuit you posted has an open circuit across the points where you measure $$\ V_c \$$ hence there is no current flowing in the circuit and therefore $$\i = 0\$$, which ultimately leads to $$\ V_c = 12 V\$$