Question

The current is suddenly turned off. How long does it take for the potential difference between points a and b to reach one-half of its initial value

Answer 1

Complete Question

The complete  question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

     The original voltage is  $V_o$

     The new voltage is $V =\frac{V_o}{2}$

     The capacitance is  $C = 150\ nF = 150 *10^{-9} \ F$

     The first resistance is  $R_i = 26 \Omega$

      The second resistance is $R_E = 200 \Omega$

Generally the equivalent resistance is  

        $R_e = R_1 + R_E$

=>     $R_e = 26 +200$

=>     $R_e = 226 \ \Omega$

Generally the time constant is mathematically represented as

     $\tau = RC$

=>  $\tau = 226 * 150 *10^{-9}$

=>  $\tau = 3.39 *10^{-5} \ s$

Generally the voltage is mathematically represented as

    $V = V_o e^{-\frac{t}{\tau} }$

=>   $\frac{V_o}{2} = V_o e^{-\frac{t}{\tau} }$

=>   $0.5 = e^{-\frac{t}{\tau} }$

=>   $ln(0.5) = {-\frac{t}{ 3.39 *10^{-5} } }$

=>  $ln(0.5) * 3.39 *10^{-5} = -t$

=>  $t = 2.35*10^{-5} \ s$