technical interview Q & A for high-tech professionals
Interview Questions
Electronics Hardware
Computer Software
Quick Thinking Tests
General questions
Phone screening questions
Submit your Q or A

Technical articles
Technical discussion
Resume and interview
How to get a job in Silicon Valley
How much are you worth on market?
Do you need an agent?

Break point
Written Exam
Logic Tests
Professional Test
Tomato company
Cup of coffee
How stock market works
Engineering jokes

About Hitequest
About Hitequest
Home page


=Electronics Hardware Questions=



Two capacitors are connected in parallel through a switch.
C1= 1uF, C2=0.25uF.
Initially the switch is open, C1 is charged to 10V. What happens if we close the switch?
No losses in the wires and capacitors.


Hint from Hitequest
The equation for charge is Q=CU
After closing the switch the charge will be distributed between the two caps.
Since there are no losses in the circuit the amount of charge remains the same:
U1C1 + U2C2 = U3(C1+C2)
Both caps will share the same voltage potential U3
U3 =  (U1C1+U2C2)/(C1+C2) = (10*1 + 0*0.25)/1+0.25 = 8
U3= 8V
What about that voltage on capacitor can not change immediately? It will be the spark and system will lose some energy.
Voltage on the cap C2 in this scenario builds up almost instantly. Why, because Tau=R*C2 and since R is assumed as zero, Tau =0s. Thus, there appears to be a sharp voltage build-up on C2.
As for the spark, some energy will be lost and the voltage on both caps be slightly less that 8V.