Physics 2025-HW5 An Investigation into RC Circuits simple relationship between t
ID: 1583856 • Letter: P
Question
Physics 2025-HW5 An Investigation into RC Circuits simple relationship between the amount of charge Q stored on the capacitor and the voltage drogp U across the capacitor C u=Q or Q.CU orCzQ As you can see from the second equation, the amount of charge that can be stored on the capacitor by looki sense, a charged capacitor can act like a battery with a positive and negative terminal. In fact, in depends on the ca pacitance the capacitor. You could also say on the voltage across ing at the first equation that a charged capacitor has a voltage across its terminals. In that some electronic circuits that use a low amount of electric power kler control systems) a capacitor is built into the circuit. It charges up during normal operating conditions and acts like a short term electric backup that is able to act like a battery if there is a brief interruption in power to the home. This way, the sprinkler system doesn't need to be reprogrammed after each brief things like microwave ovens or old TV sets are still very dangerous even if the power cable is disconnected. They can still remain charged. If you take a microwave oven apart to repair it you can even get killed by electrocution should you touch the leads of the charged capacitor. The capacitor needs to the proper experience. By the way, the capacitors we use in the lab are too small to be dangerous. power outage. You can also realize now why large capacitors in first be safely discharged, a task which should only be done by someone with Note that the units of a capacitor are Coulomb C [C]--=- VoltV Farad Because most capacitor value are very small numbers, they are usually given in units of pF (pico-Farad)-10-1 F nF (nano-Farad) 109 F F (micro-Farad)-lo-F mF (milli-Farad) 1o3 F Charging a Capacitor with a Battery Let's now see what happens when we connect a battery, resistor, and capacitor in a cireuit like this one: +5V Battery Switch 0V Capacitor Resistor 154
Explanation / Answer
Part Q6:
The voltage of the capacitor just before discharge is given by
U = Q/C
Q = 75*10-12
C = 15*10-12
So,
U = 75*10-12 / 15*10-12
Uinitial = 5 J