9 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.naturally, we assume steady state
which means the capacitor branch gets deleted
but my question is
after the branch gets deleted
Yeah, the question was ambiguous. All the solutions I saw took steady state
yeah but after steady state, why does the rightmost square of the circuit have no contribution to the current
i would've thought the net in the rightmost square would be in series with the rest of the circuit
It's connected between two zero potential points basically
Assuming negative terminal is grounded
oh
oh that makes more sense
i wonder if they specified that in the question
this is memory based
well, thanks
+solved @Opt
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