Electronics > Beginners
Can superposition principle be applied on capacitor?
orin:
It's a sneaky problem. There is an initial condition of charge on the capacitor, i.e. at t = 0, v(t) isn't 0.
At t = 0, from the v(t) given, there is 8V on the capacitor. How is that going to be taken into account?
I'd just replace the 100, 400 resistors and 12V voltage source with their Thevenin equivalent (2.4V and 80 ohms). Now you have a simple voltage divider with the 20 ohm resistor and the 80 ohm Thevenin equivalent. There is v(t) - 2.4 across the divider, so you have 0.8(v(t) - 2.4) + 2.4 across the 100 ohm resistor. Substitute in v(t) and divide by 100 to get the current through the 100 resistor.
soldar:
--- Quote from: orin on February 10, 2019, 03:09:58 am ---It's a sneaky problem. There is an initial condition of charge on the capacitor, i.e. at t = 0, v(t) isn't 0.
At t = 0, from the v(t) given, there is 8V on the capacitor. How is that going to be taken into account?
--- End quote ---
Yeah, sneaky in the sense that it is badly formulated. And, yeah, Thevenin is the way to go in any case.
emece67:
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