In a circuit with impressed voltage V (t) and inductance L, Kirchhoff's first lawdigives the relationship V = L + Ri where Ris the resistance in the circuitdtand i the current. Suppose we measure the current for several values of t andobtain:· 1-001.011-021-031.04i3:10.:3.123-143-183.24where t is measured in seconds, i in amperes, the inductance L is a constant08 henneries and the resistance is 0-142 ohms. Approximate the voltage V atthe values t = 1-00 , 1-01, 1-02, 1-03 and 1-04 using the appropriate threepoint formulas.

Given, V=Ldidt+RiL=0.98HR=0.142 ohms

Answered: In a circuit with impressed voltage V…

Similar questions

Circuit II has twice the inductance and one‑half the capacitance of Circuit I. If the resistor in Circuit II is replaced with a resistor that has twice the resistance, what is the ratio of the resonant frequencies of Circuit I to Circuit II?
The current in an Series RL circuit is described by the following differential equation:R i(t) + L di(t)/(dt) = V0where R is the resistance, L is the inductance, and V0 is the voltage of the source. a) Show that this is a separable differential equation, and then solve it. (Find i(t) in terms of R, L, V0). b) For a Series RL circuit with R = 50 (Ohm) and L = 10 (Henry) and the constant voltage V0 = 100 (Volt) applied at t = 0 by closing of a switch, find the current at t = 0.5 second.
In the analogy between an RLC circuit and a mass on a spring moving through a viscous fluid, which element in the RLC circuit represents the spring? O the resistor O the capacitor O the inductor O the power supply
1. b) Write down the second order ordinary differential equation that represents the electric circuit. Where R = 302, C=1/7 F, L = 4 H, and the voltage source E(t) = 110 Sin377t DO NOT SOLVE THE SYSTEM FOR CURRENT. R www C E(t) = Esin ot еее
The charge on the capacitor of an LC circuit with capacitance C and inductance L obeys the following equation. Find the maximum current in the circuit. q= Q cos [t/(LC)0.5] Q Q/(LC) 0.5 zero QQ²/(LC) 0.5 imax 1/(LC) 0.5
The current in an RL circuit drops from 1.4 A to 8.1 mA in the first second following removal of the battery from the circuit. If L is 8.1 H, find the resistance R in the circuit.