Question

Kirchhoft s law says that if you trace a path around any c age sources less the sum of the voltage drops must equal zero. Keeping in mind Ohm s Law, V IR, for the simple RC circuit shown in Figure 1 on the previous page where the power supply is hooked the capacitor is charging, Kirchhoff s law would say that aw says that if you trace a path around any closed-loop within a circuit, the sum of the volt- the emf - the voltage across the resistor -the voltage across the capacitor 0 where is the emf supplied by the power supply. I(r)-R is the voltage drop across the resistor, q(r) İs the charge on the capacitor at any time , and C is the capacitance of the capacitor (Note that C is a constant value for a capacitor). Note that for a capacitor in a circuit, the potential difference across the capacitor is equal to its charge divided by its capacitance, AV(0) -qC. Note that in equation (1) both the current and the charge on the capacitor are functions of time. We can substitute the fact that the definition of cur- rent is 1(1)-dy(1)/dt İnto equation (1). Equation (2) can be rearranged so that it has the form of a separable ODE with dyt) and all the terms with q) on one side, and dr and all the remaining constants on the other side. Do the algebra in the space below to rearrange and set up equation (2) as a separable ODE with dqt) and qit) on one side and dt and everything else on the other side. Be sure the dqlt) side is not multiplied by any constants.

Answer 1

〉 As per given question cwe have to he axanged the euotion1' such that it has the om of a SepaTable CDE coith dq(t) and all the forms ωǐdhq(t) One side , and dt and all Stemainigg Constant on other side 1 res dad) dat) - df So, twe can say it is in oam of ODE