RLC Circuit DC Homework Solution

[ehild]

Well $i_2=0$ since capacitor is full.There can't be current.I couldn't find $i_1$...I think its 1A but I am ot sure...I thought $ε=i_1(R_1+R_3)+L(di_1/dt)$

After very long time, the circuit is in stationary state. Nothing changes, the time derivative of the currents. i ₁=1 A and i₂=0 are correct.

Part (e) I guess Q=εC
Part (f) If its one ampere and I find charge correctly then $U_L=1/2Li^2$ and $U_C=Q^2/2C$

The capacitor is not connected directly to the battery, there are two resistors in between. Is some potential drop across those resistors?

 

[Arman777]

The capacitor is not connected directly to the battery, there are two resistors in between. Is some potential drop across those resistors?

Oh you are right yes, then $Q/C=ε-i_3R_3-i_2R_2$ $i_2=0$ and $i_3=1A$ ?

 

[ehild]

Oh you are right yes, then $Q/C=ε-i_3R_3-i_2R_2$ $i_2=0$ and $i_3=1A$ ?

Yes.

 

[Arman777]

Yes.

Thanks a Lot.

 

[ehild]

Well done!

 

[BvU]

Second well done

was what I was fishing for in #21

 

[Arman777]

Second well done

View attachment 204023

was what I was fishing for in #21

Oh well...Thanks :)