Question

A charged capacitor is connected to a resistor. After how many time constant does the energy of capacitor becomes 1/10 of its iniyial value

Answer 1

Explanation:

we have to eequate it to energy then following step

Answer 2

Answer:

Capacitor will discharge to ⅒ of energy stored in time period of 1.15τ .

Explanation:

Given that :

• Energy of capacitor becomes ⅒ of the initial value

To find :

• Time period of discharging

Solution :

• Let, a capacitor of capacitance C has initial charge Q₀.
• Let, the time constant of capacitor is τ = RC.
• The equation for discharging of a capacitor is given by

$Q = Q₀ {e}^{ - \frac{t}{τ} } \: \: - - - (1)$

• Energy stored, U in a capacitor is

$U = \frac{1}{2} \frac{ {Q}^{2} }{ C}$

• U = ⅒U (given)

$\frac{1}{2} \frac{ {Q}^{2} }{ C} = (\frac{1}{10} ) \frac{1}{2} \frac{ {Q₀}^{2} }{ C} \\ 10{Q}^{2} = {Q₀}^{2}$

• Putting value of Q in above equation, we get;

$10 {(Q₀ {e}^{ - \frac{t}{τ} } )}^{2} = {Q₀}^{2} \\ 10 {Q₀}^{2} {e}^{ - \frac{2t}{τ} }= {Q₀}^{2} \\ {e}^{ - \frac{2t}{τ} } = \frac{1}{10} \\ taking \: antilog \: on \: both \: sides \\ \frac{ - 2t}{τ} = ln \frac{1}{10} \\ \frac{ - 2t}{τ} = - ln(10) \\ 2t = ln(10)τ \\ t = \frac{2.30τ}{2} = 1.15τ$

• Hence, capacitor will discharge to ⅒ of energy stored in time period of 1.15τ .