I'm probably going to have to say C. E as it seems the steepest right around there. If I'm wrong on that, it has to be B. B
Answer:
The capacitance is cut in half.
Explanation:
The capacitance of a plate capacitor is directly proportional to the area A of the plates and inversely proportional to the distance between the plates d. So if the distance was doubled we should expect that the capacitance would be cut in half. That can be verified by the following equation that is used to compute the capacitance in such cases:
C = (\epsilon)*(A/d)
Where \epsilon is a constant that represents the characteristics for the insulator between the plates. A is the area of the plates and d is the distance between them. When we double d we have a new capacitance, given by:
C_new = (\epsilon)*(A/2d)
C_new = (1/2)*[(\epsilon)*(A/d)]
Since C = (\epsilon)*(A/d)] we have:
C_new = (1/2)*C
its c so componets arr r\very fragle im aying this because it needs more then 20 letters
Answer:
-8.56V
Explanation:
Our values are given by,
e = 6.04 V
Φ = 30.3
VC = 5.32
We can calculate the voltage across the circuit with the emf formula, that is,
Now, Using Kirchoff Voltage Law,
Finally we have the potential difference across the inductor.
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