1
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0
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Consider a simple RC circuit as shown in Figure 1.

Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V0 (i.e. charging continues for time T >> RC). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is EC.

Process 2 : In a different process the voltage is first set to $${{{V_0}} \over 3}$$ and maintained for a charging time T >> RC. Then, the voltage is raised to $${{2{V_0}} \over 3}$$ without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1.

These two processes are depicted in Figure 2.

JEE Advanced 2017 Paper 2 Offline Physics - Capacitor Question 6 English Comprehension
In Process 2, total energy dissipated across the resistance ED is
A
$${E_D} = {1 \over 3}\left( {{1 \over 2}CV_0^2} \right)$$
B
$${E_D} = 3\left( {{1 \over 2}CV_0^2} \right)$$
C
$${E_D} = 3CV_0^2$$
D
$${E_D} = {1 \over 2}CV_0^2$$
2
IIT-JEE 2012 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1
In the given circuit, a charge of $$+80$$ $$\mu C$$ is given to the upper plate of the $$4$$ $$\mu F$$ capacitor. Then in the steady state, the charge on the upper plate of the $$3$$ $$\mu F$$ capacitor is
IIT-JEE 2012 Paper 2 Offline Physics - Capacitor Question 16 English
A
$$ + \,32\,\mu C$$
B
$$ + \,40\,\mu C$$
C
$$ + \,48\,\mu C$$
D
$$ + \,80\,\mu C$$
3
IIT-JEE 2011 Paper 1 Offline
MCQ (Single Correct Answer)
+2
-0.5
A $$2$$ $$\mu F$$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $$S$$ is turned to position $$2$$ is
IIT-JEE 2011 Paper 1 Offline Physics - Capacitor Question 17 English
A
$$0\% $$
B
$$20\% $$
C
$$75\% $$
D
$$80\% $$
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