When the current in the portion of the circuit shown below is 2.40 A and increases at a rate of 0.420 A/s, the measured voltage is AVab = 8.85 V. When the current is 2.40 A and decreases at the rate of 0.420 A/s, the measured voltage is AVab = 4.90 V. Calculate the values of L and R. HINT: Use a Kirchhoff loop for each of the two scenarios and then simultaneously solve the two equations. Use the signs as given by Kirchhoff's Rules for the emf in the inductor. L = R= L. R

When the current in the portion of the circuit shown below is 2.40 A and increases at a rate of 0.420 A/s, the measured voltage is AVab = 8.85 V. When the current is 2.40 A and decreases at the rate of 0.420 A/s, the measured voltage is AVab = 4.90 V. Calculate the values of L and R. HINT: Use a Kirchhoff loop for each of the two scenarios and then simultaneously solve the two equations. Use the signs as given by Kirchhoff's Rules for the emf in the inductor. L = R= L. R

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

How much energy would you consume daily if you kept 3 lightbulbs of 25 watts on all day long?.
Consider following circuit with R1 = 60 Ω, R2 = 18 Ω, R3 = 60 / 10 Ω, R4 = 18/10 Ω and and ξ=18 V d. Write Kirchhoff’s potential difference rule for right loop e. Calculate currents i1 , i2 and i3 ? f. Calculate the potential difference of R2 ? g. Calculate the potential difference of R3 ? h. Calculate the heat dissipation of R2 resistor i. Cross sections through two long conductors of the same length and material, with square crosssections of edge lengths are shown below. Conductor Y fits snugly within conductor X, Rank theresistance of X and Y?
The current in a simple single resistor (of resistance R1) circuit is I1 = 5.1 A. A second resistor, R2 = 97 Ω, is added to the first in parallel and the current through the battery increases to I2 = 8.6 A. Input an expression for the resistance of R1.
Two resistors R1 and R2 are connected in series and parallel. The values of the resistances are:R1 = 50.0 ± 0.05 R2 = 25.0 ± 0.06 Calculate the uncertainty in the combined resistance for both the series and the parallel arrangements
In the given circuit, is changes from 8 A to 10 A at t= 0, that is, is= [ 8u(-t) + 10u() ] A. Find v) and i(). is 0.5 H The current () = ( | De t) u) A. The voltage v1 = ( )e
For the given circuit with current, voltages across elements shown below. For R = 18 Ohm, L=12.4 Henry, C=16.5 Farad and v, (t)=e3t + sin(2t) volt, Find a) It(s) V2(s) b) Vi(s) R v (t) 3 ? (1) C v,(t) e Vr. lell
In the RC circuit shown in figure 1, R₁ = 20, R₂ = 49, V = 12 V and C= 5nF. The circuit has been running for a long time with the switch (S) open. Apply Kirchhoff's Loop Rule to calculate a) the current I while the switch S is still open. b) the charge on the capacitor a long time after the switch S is closed. 2 R₂ S R₁ | | I Figure 1: An RC circuit
Consider the circuit shown below where the battery has an emf of & = 93 V, the resistors have the resistances of R1=10 0, R2 = R3 = 11 Q, and the capacitance of the capacitor is C=2 µF. The switch S is closed at t=0. The capacitor is initially uncharged. At t=0, what is the current running through R2? Please express your answer using one decimal place in units of Ampere (A). R1 S R3 R23 C
Problem2:. Consider the circuits shown below. (a) Write the equation for current using the junction rule. (b) Write two equations using loop rule (c) Use these equations to find I₁, I2, and I3? (b) What is the power dissipated across each resistor (1²R)? V₁ = 24 V 12 R₂ = 62 R₁ = 80 ½ 13 R₂=402 + ₂10 V
102 C 3. For the circuit at right, find: 102 Paralel (a) The current at pts. A, B, C, D, & E. The power dissipated by each resistor. (c) The power supplied by the battery. (d) The potential differences VB- VA, Vc- VB, VD-Vc, VE-VA. 100 V E
dv + 5v = 10B(t), dt v(0) = 0 %3D For this problem we are just going to look at the equation for the battery. A fully charged battery when connected to the circuit at time zero have a voltage source equation B(t) = 10e-0.5t %3D b. Our battery is off until one second then connects for one second, then is off for one second and does not recharge (it just retains the voltage it had when it was removed from the circuit) and then is turned on for one second, then it turned off for good. Give the function for the battery voltage in the circuit using the Heaviside function and the exponential function. B2(t) =
The capacitor in the circuit shown is fully charged by a 24 V battery. The switch is closed at t = 0. At sometime after the switch is closed, the voltage across the capacitor is measured to be 10 V. What is the current in the circuit at this time, in Ampere? C = 3.0 µF, and R = 2.0 02. Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement. Cil