JEE Advanced 2017Paper 2
- Exam
- JEE Advanced
- Year
- 2017
- Paper
- Paper 2
- Subject
- Physics
- Chapter
- Capacitance
JEE Advanced 2017 · Paper 2 · Physics · Capacitance
Paragraph: Consider a simple R C circuit as shown in Figure 1 ....
Paragraph:
Consider a simple circuit as shown in Figure .
Process 1: In the circuit the switch is closed at and the capacitor is fully charged to voltage (i.e., charging continues for time ). In the process some dissipation \left(E_{D}\right)RE_{C}\frac{V_{0}}{3}T>>R C\frac{2 V_{0}}{3}T>>R CV_{0}V_{0}2 .
Question: In Process 2, total energy dissipated across the resistance is:
Question: In Process 2, total energy dissipated across the resistance is:
Options
- A(E_D=\frac{1}{2} C V_0^2)
- B(E_D=3\left(\frac{1}{2} C V_0^2\right))
- C(E_D=\frac{1}{3}\left(\frac{1}{2} C V_0^2\right))
- D
Correct answer
C. (E_D=\frac{1}{3}\left(\frac{1}{2} C V_0^2\right))
Step-by-step solution
For process (i)
Charge on capacitor
Energy stored in capacitor
Work done by battery
Heat loss
For process (ii)
Charge on capacitor
Extra charge flow through battery
Work done by battery:
Final energy store in capacitor:
Energy store in process 2:
Heat loss in process (ii) = work done by battery in process (ii) – energy store in capacitor process (ii)
For process (iii)
Charge on capacitor
Extra charge flow through battery:
Work done by battery in this process: \left(\dfrac{C{V}_{0}}{3}\right)\left({V}_{0}\right) = \dfrac{C{V}_{0}^{2}}{3}\dfrac{1}{2}C{V}_{0}^{2}\dfrac{1}{2}C{V}_{0}^{2} - \dfrac{4C{V}_{0}^{2}}{18} = \dfrac{5C{V}_{0}^{2}}{18}\dfrac{C{V}_{0}^{2}}{3} - \dfrac{5C{V}_{0}^{2}}{18} = \dfrac{C{V}_{0}^{2}}{18}
Final energy store in capacitor:
So we can say that ${E}_{D} = \dfrac{1}{3}\left(\dfrac{1}{2}C{V}_{0}^{2}\right)
Charge on capacitor
Energy stored in capacitor
Work done by battery
Heat loss
For process (ii)
Charge on capacitor
Extra charge flow through battery
Work done by battery:
Final energy store in capacitor:
Energy store in process 2:
Heat loss in process (ii) = work done by battery in process (ii) – energy store in capacitor process (ii)
For process (iii)
Charge on capacitor
Extra charge flow through battery:
Work done by battery in this process: \left(\dfrac{C{V}_{0}}{3}\right)\left({V}_{0}\right) = \dfrac{C{V}_{0}^{2}}{3}\dfrac{1}{2}C{V}_{0}^{2}\dfrac{1}{2}C{V}_{0}^{2} - \dfrac{4C{V}_{0}^{2}}{18} = \dfrac{5C{V}_{0}^{2}}{18}\dfrac{C{V}_{0}^{2}}{3} - \dfrac{5C{V}_{0}^{2}}{18} = \dfrac{C{V}_{0}^{2}}{18}
Final energy store in capacitor:
So we can say that ${E}_{D} = \dfrac{1}{3}\left(\dfrac{1}{2}C{V}_{0}^{2}\right)