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JEE Advance - Physics (2017 - Paper 2 Offline - No. 16)

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.

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.

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.

In Process 2, total energy dissipated across the resistance ED is
$${E_D} = {1 \over 3}\left( {{1 \over 2}CV_0^2} \right)$$
$${E_D} = 3\left( {{1 \over 2}CV_0^2} \right)$$
$${E_D} = 3CV_0^2$$
$${E_D} = {1 \over 2}CV_0^2$$

Explanation

Process 1 :

Voltage is set to V0/3

Charge supplied = $${{C{V_0}} \over 3}$$

Energy supplied = $${{{V_0}} \over 3} \times {{C{V_0}} \over 3} = {{CV_0^2} \over 9}$$

Process 2 :

Voltage is raised to 2V0/3

additional charge supplied = $${{2{V_0}C} \over 3} - {{{V_0}C} \over 3} = {{C{V_0}} \over 3}$$

Energy supplied = $${{2{V_0}} \over 3} \times {{C{V_0}} \over 3} = {{2CV_0^2} \over 9}$$

Process 3 :

Voltage is raised to V0

additional charge supplied = $${V_0}C - {{2{V_0}C} \over 3} = {{C{V_0}} \over 3}$$

Energy supplied = $${V_0} \times {{C{V_0}} \over 3} = {{CV_0^2} \over 3}$$

Total energy supplied to circuit = $${{CV_0^2} \over 9} + {{2CV_0^2} \over 9} + {{CV_0^2} \over 3}$$

$$ = {6 \over 9}CV_0^2 = {2 \over 3}CV_0^2$$

Final energy stored in capacitor = $${1 \over 2}CV_0^2$$

Therefore,

Energy dissipated ED = Energy supplied $$-$$ Energy stored

$$ = {2 \over 3}CV_0^2 - {1 \over 2}CV_0^2$$

$$ = {{4CV_0^2 - 3CV_0^2} \over 6} = {1 \over 6}CV_0^2$$

$$ = {1 \over 3}\left( {{1 \over 2}CV_0^2} \right)$$

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