Didn't understand totally,
let me practice and see;
please check my calculations,
Look at new image;
New Image;
http://images.foraster.us/imgs/-647497902kondansator_anahtar.JPG
I'm trying to solve the main question, through Image 2 and Image 3...
First tell me, are both images drawed true ?
second;
Paralels has the same voltage;
The question says; Voltage of KL = V,
So, look at
IMAGE 2;
v (KL) = V --> v ( KN ) = V too
( MN ) has 2 wires, underwire and upperwire;
v ( MN_Underwire ) = v ( KN ) + v ( MK ) { Because KN and MK are Serial }
v ( MN_Underwire ) = V + v ( MK )
q= V.C -->
v ( KN ) = V and C ( KN ) =
2C ! --> q ( KN ) = V. 2C
q ( KN ) = q ( MK ) --> V . 2C = v ( MK ) . C -->
v ( MK ) = 2V !
v ( MN_Underwire ) = V + 2V =
3V
Voltage of Underwire = Voltage of Upperwire = Total Voltage ;
Voltage of System = 3V
-------------------
Now i go further to solve the problem;
IMAGE 3; (key open)
generator is same; as it is DC ...
So; System has still 3 Volt...
C_combined = 3/5 C
V2 = ?
q= V.C -->
q System = q (MK) { Because , Serial Q's are equal }
q (MK) = q (MK_Downwire) + q (MK_Upperwire)
q (MK_Upperwire) = Vupper . 0,5C
q (MK_Downwire) = Vdown . C
Vdown = Vupper ('cos parallel )
--> q (MK) = q (MK_Downwire) + q (MK_Upperwire)
--> q (MK) = Vdown.C + Vdown. 0,5C -->
q (MK) = 1,5C . Vdown
q System = q (MK)
3V . 3/5 C = Vdown . 1,5C
Vdown = Vup = 6/5 V
q = V.C --> q (ML) = q (KL) = q Upper
v ( ML ) . C =
v (KL) . C = 6/5 V . 1/2 C
v ( KL ) = 3/5 V
V2 = 3/5V = Solution
Huhh... Finally... Thanks... Check the maths all right ?:P
Any mental mistake ?
