In Process 2, total energy dissipated across the resistnace $E_D$ is:

Q: In Process 2, total energy dissipated across the resistnace \$E_D\$ is:

\$E_D= W_b- \Delta V\$\$E_D= \dfrac{CV_0}{3}[\dfrac{V_)}{3}+ \dfrac{2V_)}{3} +V_0] - \dfrac{1}{2}CV_0^2\$\$E_D= \dfrac{CV_0[2V_0]}{3} - \dfrac{1}{2}CV_0^2= \dfrac{CV_0^2}{6}\$Hence D is correct answer.

Question

Accepted Answer

$E_D= W_b- \Delta V$$E_D= \dfrac{CV_0}{3}[\dfrac{V_)}{3}+ \dfrac{2V_)}{3} +V_0] - \dfrac{1}{2}CV_0^2$$E_D= \dfrac{CV_0[2V_0]}{3} - \dfrac{1}{2}CV_0^2= \dfrac{CV_0^2}{6}$Hence D is correct answer.

Electrostatic Potential And Capacitance

Physics
1435 Views

In Process 2, total energy dissipated across the resistnace $E_{D}$ is:

Electrostatic Potential And Capacitance

Physics 1435 Views

In Process 2, total energy dissipated across the resistnace ED​ is:

Physics
1435 Views

In Process 2, total energy dissipated across the resistnace $E_{D}$ is: