loadpowerresistanceresistors
I'm trying to solve a homework problem to find the power delivered to the 10Ω load.
So far I found the total resistance, which is 12Ω.
And I think I have the right formula to work out the power, correct me if there's another way or if I'm wrong.
My calculations so far
My next calculations
Any help will be greatly appreciated. 🙂
One of the benefits of the hydro power plant is that it can be throttled all the way down to zero. Hydro power plants are good at supplying peaky loads. If a standalone hydro power plant has spare capacity, it just throttles down and conserves the water, rather than pumping it back up into the reservoir. These considerations make your problem (in its original form) look less intuitive and realistic.
However, your problem will look more intuitive and realistic if you imagine that the nominal power Pnom = 10 GW is coming from an external supply other than the hydroelectric plant. This external supply can not throttle up or down. Also, there is a local hydroelectric facility, which can pump the reservoir when Pload < Pnom and use it to augment the power supply when Pload > Pnom. This scheme can support Pload,max > Pnom.
At τsw the hydroelectric system switches from pumping storage to using it. $$P_{nom}=P_{load,max} \cdot cos(\pi \tau_{sw}/24)$$
Energy is in [GW⋅h] units. Energy balance of the pumped storage. $$E_{storage,out}=0.7 E_{storage,in}$$
Substitute the appropriate integrals for the energies, and you get a system of two equations with two variables τsw and Pload,max.
This is a rather poorly set question- no information is given as to the initial condition of the capacitor. If the 10uF cap happens to be charged to 12V the current will be 0 before and after the switch closes.
Anyway, what they want you to do is to assume the initial voltage on the capacitor is 0V.
You should then be able to write down the equation for voltage on the capacitor as a function of time (memorize this- it's the solution to the differential equation if you want to do it from first principles). $$ V_c(t) = 12 (1 - e^{-t/\tau}) $$ where \$\tau = RC\$.
Since you know that the current is \$(12V - V_c(t)) \cdot 2.2k\$ you can solve for when current is 2mA.
Alternately you can recognize that $$i(t) = I_0 e^{-t/\tau}$$ where \$I_0 = 12V/2.2K\$ (you have this)
so $$ t = -\text{ln}(2mA/I_0)* RC $$ and the answer is 'b' 22ms
To get the above line, divide both sides by \$I_0\$, then take the ln() of both sides and solve for t.
Best Answer
Do Ohm's law to find \$I_T\$.
Work back through your circuits. The hardest part is getting students to do sketches. It helps you reduce the circuit to find equivalent resistance AND to calculate branch currents and voltages. You have so that allows you to work backwards.
If series, calculate voltages. Then if resistors are in parallel, you will know voltages. If parallel, calculate branch currents. If branch has series resistors, you will know current.
With the circuit don't focus on how it looks, focus on how it is connected. \$R_8\$ and \$R_6\$ are in parallel. Key off your resistance calculations.
In series, current is constant, work out voltages with Ohm's Law.
In parallel, voltage is constant, work out branch currents with Ohm's Law. If resistors are equal, current splits equally.
A trick I used was to take all outside resistors and add up their voltages. If it equals \$V_T\$, then you are correct. Basic Kirchhof's Voltage Law (KVL).
In your case: $$V_T = V_1 + V_2 + V_3 + V_4 = 24V$$
Any path for current, must satisfy KVL. But it's easiest to take the outside resistors.