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- Two capacitors are connected in parallel across the terminals of a battery. One has a capacitance of 2.5 μF and the other a capacitance of 5.4 μF. These two capacitors together store 7.1 x 10-5 C of charge. What is the voltage of the battery?Four uncharged capacitors with equal capacitances are combined in series. The combination is connected to a 7.65 V battery, which charges the capacitors. The charging process involves 0.000193 C of charge moving through the battery. Find the capacitance C of each capacitor. C = FA system of capacitors is shown below. What is the equivalent capacitance of the system if the individual capacitors are: Express the answer in microFarads. For example, if the answer came out to be 4.98 x 10-7 F, this is the same as 49.8 microFarads. You would then answer with numeric value of 4.98. C1 = 6.6 x 10-6 F C2 = 5.9 x 10-6 F C3 = 3.4 x 10-6 F C4 = 4.4 x 10-6 F C5 = 8.1 x 10-6 F
- 28. Write the node voltage equations for Figure 9-29. Use your calculator to find the node voltages.A system of capacitors is shown below. What is the equivalent capacitance of the system is the individual capacitors are; Express the answer in microFarads. For example, if the answer came out to be 4.98 X 10-7 F, this is the same as 49.8 microFarads. You would then answer with the numeric value of 4.98. C1 = 5.5 x 10-6 F C2 = 5.7 x 10-6 F C3 = 8 x 10-6 F C4 = 4.5 x 10-6 F C5 = 2.1 x 10-6 FA person stands directly in front of two speakers that are emitting the same pure tone. The person moves to one side until no sound is heard. At that point, the person is 7 m from one speaker and 7.2 m from the other. What is the frequency of the tone being emitted? Take the speed of sound to be 344 m/s, as usual.
- A system of capacitors is shown below. What is the equivalent capacitance of the system is the individual capacitors are; Express the answer in microFarads. For example, if the answer came out to be 4.98 X 10-7 F, this is the same as 49.8 microFarads. You would then answer with the numeric value of 4.98. C1 = 6.7 x 10-6 F C2 = 4.2 x 10-6 F C3 = 2.1 x 10-6 F C4 = 2.6 x 10-6 F C5 = 6.3 x 10-6 FHow would you rearrange that equation for m_0It takes 112 J of work to move 2.3 C of charge from the negative plate to the positive plate of a parallel plate capacitor. What voltage difference exists between the plates? Answer in units of V.
- A student is trying to build a home-made capacitor. He uses two 1.0 cm x 1.0 cm aluminum foil as the two parallel conducting plates, spaced 0.40 cm apart. He then connects the two foils with a 3.0 V battery. How many electrons are stored in the negative plate?How much work does the battery do in moving 5 μ C from its low terminal to its high terminal? Answer in μJ. Diagram is attached.How do I rearrange this equation so that I am solving for t?