Answer:
The time it will take the capacitor to loss half of its stored energy is 11.1 s.
Explanation:
Given;
capacitance, C = 4.0-mF = 0.004 F
resistance, R = 4.0-kΩ = 4000 Ω
The voltage of a capacitor at an instant time is given by;
[tex]V = (V_o)e^{-t/RC}[/tex]
the final voltage, V is half of the initial voltage, V₀ at the time t
V = V₀/2
[tex]V = (V_o)e^{-t/RC}\\\\\frac{V}{2} = (V_o)e^{-t/RC}\\\\0.5 = e^{-t/RC}\\\\ln(0.5)= ln(e^{-t/RC})\\\\ln(0.5)= \frac{-t}{RC}\\\\ t = -(RC)ln0.5\\\\t = -(4000*0.004)ln(0.5)\\\\t = 11.1 \ s[/tex]
Therefore, the time it will take the capacitor to loss half of its stored energy is 11.1 s.
The time taken by capacitor will be "11.1 s",
According to the question,
Capacitance, C = 4.0 mF or,
= 0.004 F
Resistance, R = 4.0 kΩ or,
= 4000 Ω
As we know the relation,
→ V = (V₀)[tex]e^{-\frac{t}{RC} }[/tex]
or,
[tex]\frac{V}{2}[/tex] = (V₀)[tex]e^{-\frac{t}{RC} }[/tex]
0.5 = [tex]e^{-\frac{t}{RC} }[/tex]
By taking "log" both sides, we get
ln(0.5) = ln([tex]e^{-\frac{t}{RC} }[/tex])
Now,
The time will be:
t = -(RC)ln(0.5)
By substituting the values,
= -(4000 × 0.004)ln(0.5)
= 11.1 s
Thus the above approach is right.
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