Need some coaching,not sure where to start (KVL)

[Grunt]

Trying to get back back into this. been away for a very long time. Taking a course on the fundamentals of basic electronics.
Here is what I need help with:


The following circuit depicts a blocking oscillator during the initial instant after triggering. By means of Kirchhoff’s Voltages Law, find the currents flowing in the branches containing the 93.6 and 164 volt batteries. Also calculate the value of ec.

 

[Colin Mitchell]

This is not a basic electronics question.

 

[Grunt]

This is not a basic electronics question.

Well, its in my "Fundamentals of Basic Electronics" student guide.
Can you help ?

 

[Laplace]

The usual way to apply KVL is by using mesh currents; however, since branch currents have already been identified then one can use those instead. KVL states that the sum of voltage drops around a loop equals zero. There are two loops here so there must be two loop equations: {clockwise}

(Vbd-Vce)+(Vce-Va)+(Va-Vbd)=0 {constraint: Vbd-Va=300}

(Vd-Ve)+(Vc-Vb)=0 {constraint: Vc=Ve, Vb=Vd}

The next step is to expand the loop equations in terms of the branch currents and the known voltages. Then solve the equations for the branch currents I' & I". You'll have to decide what it means to "calculate the value of ec" - I have no idea.

 

[Colin Mitchell]

Start by removing the 5v and put 295v as the supply.
Now we have a 211 ohm resistor with 47.5mA flowing through it. Work out the voltage that will be developed this resistor.
Now take this voltage off the supply voltage of 295v = X
For the left-hand path we have 21.4k plus 9.74k in series and the effective voltage is X - 93.6v.
You can now work out the current-flow.
For the right-hand path it is X - 164v and a resistance of 2.92k.

 

[Grunt]

Start by removing the 5v and put 295v as the supply.
Now we have a 211 ohm resistor with 47.5mA flowing through it. Work out the voltage that will be developed this resistor.
Now take this voltage off the supply voltage of 295v = X
For the left-hand path we have 21.4k plus 9.74k in series and the effective voltage is X - 93.6v.
You can now work out the current-flow.
For the right-hand path it is X - 164v and a resistance of 2.92k.

Thank You,
I will work it later and see if I acheive the correct answer.

 

[Grunt]

The usual way to apply KVL is by using mesh currents; however, since branch currents have already been identified then one can use those instead. KVL states that the sum of voltage drops around a loop equals zero. There are two loops here so there must be two loop equations: {clockwise}

(Vbd-Vce)+(Vce-Va)+(Va-Vbd)=0 {constraint: Vbd-Va=300}

(Vd-Ve)+(Vc-Vb)=0 {constraint: Vc=Ve, Vb=Vd}

The next step is to expand the loop equations in terms of the branch currents and the known voltages. Then solve the equations for the branch currents I' & I". You'll have to decide what it means to "calculate the value of ec" - I have no idea.

Thank You