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

Grunt

May 19, 2015
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May 19, 2015
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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.

upload_2015-5-19_9-15-8.png
 

Laplace

Apr 4, 2010
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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

Aug 31, 2014
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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

May 19, 2015
4
Joined
May 19, 2015
Messages
4
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

May 19, 2015
4
Joined
May 19, 2015
Messages
4
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
 
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