Hey everyone, I have attached the question and solution to this problem but do not arrive at the same solution. My problem is that I do not know how to use the \(Ib_2\) current so my approach was just to say that another current would flow through the capacitor and that is equal to the current flowing through the resistor:
\(I_r_e_s = I_c_a_p\)
\( \frac{v_i - 0}{R} = -c\frac{dV_o}{dt}\)
\(-\frac{v_o}{RC} = \frac{dv_o}{dt}\)
\(V_o(t) = -\frac{1}{RC} \int^t_0 V_i(t) + V_o(0)\)
so \(V_o(t) = -\frac{1}{RC} \int^t_0 V_i(t) \)
Any advice why the above is not applicable to this question and how I should go about it? Thanks!
\(I_r_e_s = I_c_a_p\)
\( \frac{v_i - 0}{R} = -c\frac{dV_o}{dt}\)
\(-\frac{v_o}{RC} = \frac{dv_o}{dt}\)
\(V_o(t) = -\frac{1}{RC} \int^t_0 V_i(t) + V_o(0)\)
so \(V_o(t) = -\frac{1}{RC} \int^t_0 V_i(t) \)
Any advice why the above is not applicable to this question and how I should go about it? Thanks!
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