Proceeding with the integral, which takes a quadratic form in q, gives a summed energy on the capacitor Q 2/2C = CV b 2/2 = QV b/2 where the V b here is the battery voltage. Part of the intuitive part that goes into setting up the integral is that getting the first element of charge dq onto the capacitor plates takes much less work because most of the battery voltage is dropping across the resistance R and only a tiny energy dU = dqV is stored on the capacitor.
It's not at all intuitive in this exponential charging process that you will still lose half the energy into heat, so this classic problem becomes an excellent example of the value of calculus and the integral as an engineering tool. I'm just going to lower the resistance of the charging pathway so I will get more energy on the capacitor." This doesn't work, because the energy loss rate in the resistance I 2R increases dramatically, even though you do charge the capacitor more rapidly. The counter-intuitive part starts when you say "That's too much loss to tolerate. But half of that energy is dissipated in heat in the resistance of the charging pathway, and only QV b/2 is finally stored on the capacitor at equilibrium. To be sure, the battery puts out energy QV b in the process of charging the capacitor to equilibrium at battery voltage V b. The problem of the " energy stored on a capacitor" is a classic one because it has some counterintuitive elements. Where did half of the capacitor charging energy go? The form of the integral shown above is a polynomial integral and is a good example of the power of integration.
Summing all these amounts of work until the total charge is reached is an infinite sum, the type of task an integral is essential for.
Transporting differential charge dq to the plate of the capacitor requires workīut as the voltage rises toward the battery voltage in the process of storing energy, each successive dq requires more work. Note that the total energystored QV/2 is exactly half ofthe energy QV which is supplied by the battery, independent of R! Summing these continuously changing quantities requires an integral. As the charge builds up in the charging process, each successive element of charge dq requires more work to force it onto the positive plate. Storing energy on the capacitor involves doing work to transport charge from one plate of the capacitor to the other against the electrical forces. The voltage on the capacitor is proportional to the charge When the switch is closed to connect the battery to the capacitor, there is zero voltage across the capacitor since it has no charge buildup. HyperPhysics***** Electricity and Magnetism
But the battery energy output is QV! Where did half of the energy go? This energy expression can be put in three equivalent forms by just permutations based on the definition of capacitance C=Q/V. If Q is the amount of charge stored when the whole battery voltage appears across the capacitor, then the stored energy is obtained from the integral: The voltage V is proportional to the amount of charge which is already on the capacitor. Voltage represents energy per unit charge, so the work to move a charge element dq from the negative plate to the positive plate is equal to V dq, where V is the voltage on the capacitor. The energy stored on a capacitor can be expressed in terms of the work done by the battery. Energy Stored on a Capacitor Storing Energy in a Capacitor
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