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
The new natural frequency would be ω/2.
we know that,
= ω. -> equation 1
now, when capacitance is quadrupled,
. -> equation 2
substituting value of equation 1 in equation 2 , we get,
Hence, the new natural frequency of the circuit is ω/2.
what do you mean by frequency ?
The resonant frequency for a particular circuit is the frequency at which this equality stands true. Where L is the inductance in henries and C is the capacitance in farads, this is the LC circuit's resonant frequency.
Learn more about frequency here:-
brainly.com/question/12530980
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Answer:
V = 50 volts
Explanation:
Given that,
Resistance, R = 10 ohms
Current, I = 5 A
We need to find the potential difference across the circuit. We know that,
V = IR
Put all the values,
V = 5 × 10
V = 50 volts
Hence, the potential difference is equal to 50 volts.
