A displacement current of 4.0 text{ A} can be | vedclass

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(B) The displacement current $I_d$ in a capacitor is given by the formula $I_d = C \frac{dV}{dt}$,where $C$ is the capacitance and $\frac{dV}{dt}$ is

Vedclass · Accepted Answer

$0.67$

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(B) The displacement current $I_d$ in a capacitor is given by the formula $I_d = C \frac{dV}{dt}$,where $C$ is the capacitance and $\frac{dV}{dt}$ is the rate of change of potential difference.Given values are $I_d = 4.0 	ext{ A}$ and $C = 6 	ext{ }\mu	ext{F} = 6 	imes 10^{-6} 	ext{ F}$.Rearranging the formula to find the rate of change of potential difference: $\frac{dV}{dt} = \frac{I_d}{C}$.Substituting the values: $\frac{dV}{dt} = \frac{4.0}{6 	imes 10^{-6}} = \frac{4.0}{6} 	imes 10^6 	ext{ V/s}$.Calculating the value: $\frac{dV}{dt} = 0.666... 	imes 10^6 	ext{ V/s}$.Rounding to two decimal places,we get $\frac{dV}{dt} \approx 0.67 	imes 10^6 	ext{ V/s}$.Comparing this with $\alpha 	imes 10^6 	ext{ V/s}$,we find $\alpha = 0.67$.

$A$ displacement current of $4.0 \text{ A}$ can be set up in the space between two parallel plates of a $6 \text{ }\mu\text{F}$ capacitor. The rate of change of potential difference across the plates of the capacitor is nearly $\alpha \times 10^6 \text{ V/s}$. The value of $\alpha$ is . . . . . . .

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