An R-C Circuit 9 of 15 Review | Constants Part C Learning Goal: Now solve the differential equation V(t) = -CR- for the initial conditions given in the problem introduction to find the voltage as a function of time for any time t. To understand the behavior of the current and voltage in a simple R-C circuit Express your answer in terms of go , C, R, and t. A capacitor with capacitance C is initially charged with charge go . At time t = 0 a resistor with resistance R is connected across the capacitor. (Figure 1) View Available Hint(s) V (t) = 10e Rot Submit Previous Answers Correct If there were a battery in the circuit with EMF &, the equation for V (t) would be V(t) = & - RC -. This differential equation is no longer homogeneous in V (t) (homogeneous means that if you multiply any solution by a constant it is still a solution). However, it can be solved simply by the substitution Vb (t) = V(t) - &. The effect of this substitution is to eliminate the & term and yield an equation for Vi (t) that is identical to the equation you solved for V (t) . If a battery is added, the initial condition is usually that the capacitor has zero charge at time t = 0. The solution under these conditions will look like V(t) = E(1 - e-/(RC) ). This solution implies that the voltage across the capacitor is zero at time t = 0 (since the capacitor was uncharged then) and rises asymptotically to & (with the result that current essentially stops flowing through the circuit). Part D Figure Given that the voltage across the capacitor as a function of time is V(t) = et/() , what is the current I (t) flowing through the resistor as a function of time (for t > 0)? It might be helpful to look again at Part A of this problem. Express your answer in terms of t and any quantities given in the problem introduction. View Available Hint(s) I(t) JAL AEd 0 2 ? + 90 +++t I (t) = Submit Provide Feedback Next >