Answer:
The answer is "0.047".
Explanation:
Given value:
[tex]\to t_{\frac{1}{2}}= 0.4406\\\\\to V=0.9375 V_{max}[/tex]
Calculating the capacitance:
[tex]\to V=V_{max} (1-e^{\frac{-t}{Rc}})[/tex]
In this, the t = time, which is taken to calculates its maximum voltage.
[tex]\to 0.9375 V_{max} = V_{max}(1-e^{- \frac{t}{103\times(165.279\times10^{-6})}})\\\\\to e^{- \frac{t}{103\times(165.279\times10^{-6})}}= 0.0625\\\\\to - \frac{t}{103\times(165.279\times10^{-6})} = \ln(0.0625)\\\\\to -t= 0.01702\times(-2.77258) \\\\ \to -t = -0.04719 \\\\ \to t= 0.04719 \approx 0.047 \ s[/tex]
