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(Figure 1) shows the voltage across a 0.10 μF capacitor. Select the correct graph showing the displacement current through the capacitor as a function of time.

Figure 1 Shows The Voltage Across A 010 ΜF Capacitor Select The Correct Graph Showing The Displacement Current Through The Capacitor As A Function Of Time class=
Figure 1 Shows The Voltage Across A 010 ΜF Capacitor Select The Correct Graph Showing The Displacement Current Through The Capacitor As A Function Of Time class=
Figure 1 Shows The Voltage Across A 010 ΜF Capacitor Select The Correct Graph Showing The Displacement Current Through The Capacitor As A Function Of Time class=
Figure 1 Shows The Voltage Across A 010 ΜF Capacitor Select The Correct Graph Showing The Displacement Current Through The Capacitor As A Function Of Time class=
Figure 1 Shows The Voltage Across A 010 ΜF Capacitor Select The Correct Graph Showing The Displacement Current Through The Capacitor As A Function Of Time class=

Sagot :

Answer:

The answer is the third photo

Explanation:

View image bigdmarx2001

Following are the calculation to the voltage:

Given:

The voltage across a [tex]\bold{0.10 \mu \ F}[/tex] capacitor.

To find:

The function of the capacitor=?

Solution:

[tex]\to \bold{Q = CV}[/tex]

so,  

[tex]\to \bold{I = \frac{dQ}{dt} = C\ \frac{dV}{dt}}[/tex]

As a result, this should nearly correspond to the gradient of the voltage curve.

from 0 to 1:

[tex]\to \frac{dV}{dt} = \frac{100}{1 \mu \ S}[/tex]

so  [tex]I = 0.1 \times 100 = 10\ A[/tex]

from 1 to 2:

[tex]\to \frac{dV}{dt} = 0[/tex]

so [tex]I = 0[/tex]

from 2 to 4:

[tex]\to \frac{dV}{dt} = \frac{100}{2 \mu \ S}[/tex]

so [tex]I = 0.1 \times 50 = -5 A[/tex]

Therefore, the final answer is "third image".

Learn more:

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