In order to establish an instantaneous displacement current of 1 mA in the space between the plates of 2 μ F parallel plate capacitor, the rate of change of potential difference is

Question

In order to establish an instantaneous displacement current of 1 mA in the space between the plates of 2 μ F parallel plate capacitor, the rate of change of potential difference is (A) 100 V s-1 (B) 200 V s-1 (C) 300 V s-1 (D) 500 V s-1

Tardigrade Admin · Accepted Answer

ID= 1 mA = 10-3 A; C = 2 μ F = 2 × 10-6 F ID = IC = (d/dt)(CV) = C (dV/dt) Therefore, (dV/dt) = (ID/C) = (10-3/2 × 10-6) = 500 V s-1 Therefore, applying a varying potential difference of 500 V s-1 would produce a displacement current of desired value.

Q. In order to establish an instantaneous displacement current of $1 m A$ in the space between the plates of $2 μ F$ parallel plate capacitor, the rate of change of potential difference is

$100 V s^{- 1}$

$200 V s^{- 1}$

$300 V s^{- 1}$

$500 V s^{- 1}$

Q. In order to establish an instantaneous displacement current of 1mA in the space between the plates of 2μF parallel plate capacitor, the rate of change of potential difference is

100Vs−1

200Vs−1

300Vs−1

500Vs−1

ID​=1mA=10−3A; C=2μF=2×10−6F ID​=IC​=dtd​(CV)=CdtdV​ Therefore, dtdV​=CID​​=2×10−610−3​=500Vs−1 Therefore, applying a varying potential difference of 500Vs−1 would produce a displacement current of desired value.

Q. In order to establish an instantaneous displacement current of $1 m A$ in the space between the plates of $2 μ F$ parallel plate capacitor, the rate of change of potential difference is

$100 V s^{- 1}$

$200 V s^{- 1}$

$300 V s^{- 1}$

$500 V s^{- 1}$