The current in an RL circuit decreases from 0.50 A to 5.0 mA in the one second after the battery is switched out of the circuit. If L = 11.4 H, what is the resistance in the circuit? Express your answer in ohms.
let T is the time constant of the circuit.
use,
I(t) = I_max*e^(-t/T)
5*10^-3 = 0.5*e^(-1/T)
10^-2 = e^(-t/T)
ln(10^-2) = -t/T
==> T = -t/ln(10^-2)
= -1/ln(10^-2)
= 0.2171 s
we know, T = L/R
==> R = L/T
= 11.4/0.2171
= 52.5 ohms <<<<<<<<<<<<-----------------------Answer
The current in an RL circuit decreases from 0.50 A to 5.0 mA in the one...
The current in an RL circuit drops from 0.77 A to 13 mA in the first second following removal of the battery from the circuit. If L is 13 H, find the resistance R in the circuit.
The current in an RL circuit drops from 1.70 A to 33.0 mA in the first second following removal of the battery from the circuit. Find the inductive time constant. If A is 24.0 H, find the resistance R in the circuit.
The current in an RL circuit drops from 0.90 A to 11 mA in the first second following removal of the battery from the circuit. If Lis 11 H, find the resistance R in the circuit. Number Units
For the RL circuit pictured below, L = 86.5 Henries and R = 8.5 Ohms. The switch is placed in position 1 for a long time before it is switched to position 2. After waiting 3 seconds, the current in the circuit is 8.9 Amperes. What is the potential difference across the battery?
For the RL circuit pictured below, L = 86.5 Henries and R = 8.5 Ohms. The switch is placed in position 1 for a long time before it is switched to position 2. After waiting 3 seconds, the current in the circuit is 8.9 Amperes. What is the potential difference across the battery?
ON CAPAC Transient RL Circuit Consider the circuit shown in the figure. The resistance of the wire used to make the inductor is negligible compared to the resistors in the circuit. V-140 V, R; -10.Ohms, Ry - 1100 Ohms, and L-200 H RI WWW For this part, assume that switch S has been closed for a long time so that steady currents exist in the circuit. Find (1) the battery current, (2) the current in resistor Ry, and (3) the...
4. The current in an RL circuit increases to 95% of its final value 3 seconds after the switch is closed. The inductance in the circuit is 0.310 H. Find the inductive time constant and resistance for the circuit. (Assume the inductor has neglible resistance). Show your work. 5. At resonance, the reactance of the inductor (L) and the capacitor (C) cancel each other so that the impedance (Z) is just the resistance (R). Using the values from this experiment...
In a RL series circuit, the current reaches half its maximum value in 2.0 s after the switch is closed. If the resistance is 5.0 Ω, find the inductance of the inductor. (4 pts) In a RL series circuit, the current reaches half its maximum value in 2.0 s after the switch is closed. 5) If the resistance is 5.0 Ω, find the inductance of the inductor. (4 pts)
E S Figure 31.2: RL Circuit (Problems 8 and 10) 10. In the circuit given in Fig. 31.2, R 20 kS2, R220 k2, L 50 mH, and the ideal battery has E40 V. The switch has been open for a (a) Just after the switch was closed, what are the current through the (b) After a very long time, what is the current through the battery (c) What is the current through the battery and its rate of change at...
RL Circuits The current in the RL circuit shown below reaches half its maximum value in 1.75 ms after the switch S1 is thrown. Determine (a) the time constant of the circuit and (b) the resistance of the circuit if L = 250 mH.