A circuit (shown to the right) initially consists of a parallel plate capacitor
ID: 1602376 • Letter: A
Question
A circuit (shown to the right) initially consists of a parallel plate capacitor connected to the terminals of a 5.0 V battery. For each of the changes to the circuit detailed below, determine whether the energy stored by the capacitor between points D and E will increase, decrease, or remain unchanged.
Sapling Learning A circuit (shown to the right) initially consists of a parallel plate capacitor connected to the terminals of a 5.0 V battery. For each of the changes to the circuit detailed below, determine whether the energy stored by the capacitor between points Dand E will increase, decrease, or remain unchanged Incorrect. The positive terminal of a battery is at higher potential than the negative terminal Unchanged Decreases ncreases The wire from A to B is replaced by Another capacitor is another capacitor connected between The wire connecting points points D and A C and D is replaced with a 2.0 V battery, with the terminal connected to C The terminals of The area of the the 5.0 V battery capacitors plates are are reversed quadrupled, while its plate separation is doubled MapExplanation / Answer
1.
*In Decrease :
energy stored is given as then
E = (0.5) CV2
where V = Voltage across the capacitor's plates
as we add one more capacitor in series , the battery Voltage gets divided. hence Capacitor in discussion will have less Voltage than earlier. hence the energy stored will decrease.
2.
*Reversing the battery terminal will not change the Voltage across the capacitor and its capacitance. hence Energy stored will remain same .
3.
*as the new battery is connected with opposite polarity , the overall voltage across the capacitor becomes
5 - 2 = 3 volts
hence the energy stored will decrease.
4.
*In as the area is quadrupled and plates saperation is doubled , the capacitance becomes twice its previous value. hence the energy stored will increase as C has increased.
5.
*Decrease :
energy stored is given as then
In E = (0.5) CV2
where V = Voltage across the capacitor's plates
as we add one more capacitor in series , the battery Voltage gets divided. hence Capacitor in discussion will have less Voltage than earlier. hence the energy stored will decrease.
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