This helped me so much! Before watching, I kept forgetting that the problem I am working on was under steady-state conditions. Your video cleared that mistake up and a few other concepts as well! Thanks again!
Thanks for this great explanation you took the Req= 6 then you separated them again when u performed kvl to find vc . Why cant we perform kvl with Req to get vc ?
Just out of curiosity.When finding the Voltage across the capacitor, why didn't you take the KVL around the first loop? does it matter which loop you use?why the second loop?
+chisom nnoka No particular reason -- that's just what I felt like at the time. You could do the same with a KVL around the first loop (12V source - 2V drop across 1ohm = 10V across cap). With circuits, there are often multiple ways to solve the same problem.
It doesn't matter. If you apply KVL to left loop in clockwise direction, we get that -12 V + (1 Ω) _i_ + (4 Ω)(0 A) + _v_C_ = 0 V -12 V + (1 Ω)(2 A) + 0 V + _v_C_ = 0 V -12 V + 2 V + _v_C_ = 0 V -10 V + _v_C_ = 0 V _v_C_ = 10 V which is the same. This is basic knowledge; you should keep studying the basics before studying first- and second-order DC circuits.
Dr. Montgomery Thanks for the video. I have one question regarding the voltage conservation concept... the total voltage source is 12. Vc is 10 and the voltage for 1 Ohms is (1x2=2 v), for the 5 Ohm's resistor is 5*2= 10 volt. The total voltage is 10+2+10= 22 volt. I'm a little bit confused since the total voltage drop is greater than the voltage drop provided by the source.
In parallel configurations, the voltage is split. Run a KVL around the entire circuit (you can ignore the middle branch because it is an open circuit), and you will find that the total voltage is 12V. If you were to include the middle branch, you would still get 12V because you subtract the voltage flowing against the resistor in both KVL equations which will cancel the additional 10 out. I hope that makes sense.
please give me solution urgent ... Derive the analytical expression of the capacitance , c ,of two parallel plates having a dielectric material in between them of relative permittivity of €r. Find the maximum voltage that can be applied across a parallel plate capacitor of 0.006uF having a plate area of 0.02m2. The dielectric is porcelain. Assume a linear relationship between the dielectric strength and the thickness of the dielectric . For porcelain , dielectric strength =200 Volts /Mil and €r=6.0.
When you worked out Vc = 10V Wouldn't the voltage across the capacitor be 2 volts? The 10 volts is the voltage drop across the two resistors and then 2 volts to be lost to the capacitor since we started with 12?
+VT Gaming 10V is the drop across the 5ohm resistor, which is in parallel across the capacitor, so the voltage across the capacitor must be the same. Recall there is no voltage drop across the 4ohm resistor because there is no current flow through it at steady state. The remaining 2V from the supply is dropped across the 1ohm resistor. The result is the same if you calculate using a voltage divider expression (as opposed to using Ohm's Law as I show).
Hello sir. I have a genuine question. I'm a new student. If the two leads of a resistor are A and B. If one connects the +ve terminal of an ideal 5V voltage source to A and the -ve terminal to lead B and then connect only the positive terminal of a 7V ideal voltage source to lead A, what would the potential be at the lead A. It would be nice if you could suggest a book on basic circuit elements. Thanks.
I'm not quite sure I understand the situation...if only 1 terminal of the 7V source is connected to the resistor, then it has no impact on the voltage across the resistor. You would only have the 5V across the resistor. The book I used for this course was "Electric Circuits" 10th or 9th ed, by Nilsson & Riedel. But, there are many good books on circuits out there...
There can't be current flowing through the 4 ohms resistor because the capacitor is blocking the current as we are dealing with DC current. So the resistor never makes his way to the negative pole of the battery.
please I need u to derive the resonant frequency when it involve R1, l1 in parallel, R2, c2 in parallel, L2, C2 in parallel and all the circuit element connected in parallel
I think no , cause for example power of inductor= v*i =L(di/dt)*i so ( dc means no change in current so means no power is taken or dissipated through dc so power =0 and energy = integration of power over time so it equals 0 ) .
How would we find the energy supplied by the 12V battery? And if inductors and capacitors store energy then what about resistors, don't they dissipate energy? So what will the total energy be- This is unclear to me.
The general equation to find the energy in an electric component (whether it's DC, AC, transient state, steady state, etc.) is _w(t)_ = ∫ _p(t) dt_ + _W_0._ Since this is DC and in steady state, _v(t)_ and _i(t)_ are constant, so _p(t)_ is constant too. Therefore, if we also assume the initial energy is 0 J, then the previous equation simplifies to _w(t)_ = p(t) ∫ _dt_ or _W_ = _P_ Δt, where Δt is the time interval (notice I've used capital letters, which indicate constant values.) As you can see, the previous equation _requieres_ to specify the interval of time Δt in which we want to determine the energy. Since we haven't defined any time, we can't compute the energy in resistors nor in the voltage source. That *doesn't* mean they aren't absorbing/generating power. You might ask if that's the case, then why can we compute the energy in capacitors and inductors but not in resistors and sources. This is because that energy represents the energy *stored* by the capacitor and by the inductor, which *doesn't* depend of time since the circuit has reached steady state (and so the stored energy won't increase.) In resistors, they _always_ (even in AC) absorb power (use energy) and they don't store it, so calcuting the energy without specifying a time interval doesn't make sense (what energy does it represent? Stored? It's zero. Used? We'd need to define a time interval.) Similarly in the case of sources.
An inductor in a steady state is an "ideal inductive load" and draws no power. Also at the end of the day, an inductor is just a wire loop, so it lets everything flow through it, and as it draws no power theoretically the resistance is zero, which means current can go through it, just like as if it were a straight wire. Hence we replace it with a wire calling it a short. Hope you understand.
this is stupid because the only real information that was given about inductors and capacitors is that it will become open and short circuits after some time and the rest is basic circuit analysis
