You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an RC circuit with a half-life of t₁/2 = 3.00 s. Your supervisor is concerned that the initiation of the process is occurring too soon and that the half- life needs to be extended. He asks you to change the resistance of the circuit to make the half-life longer. All you can find in the supply room is a single 35.00 resistor. You look at the RC circuit and s e that the resistance is 26.0 0. You combine the new resistor with the old to extend the half-life of the circuit. Determine the new half-life (in s). s
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- A cell of em.f. 2V and external resistance 0.5 Q is connected across a resistor (R). The current that flows is same as that, when a cell of e.m.f. 1.5 V and external resistance 0.3 Q is connected across the same resistor. Then R=03Q (b) R=0.60 R=05Q (d) R=0.75 92A TV works properly under the voltage of ΔV = 180 V with power P = 1740 W. (a) Express the current I through the power P and the voltage ΔV. (b) Calculate the working current of the TV in A. (c) Express the resistance R through the voltage ΔV and the current I. (d) Calculate the numerical value of R in Ω.E10P2
- An initially uncharged 4.47 x 10-6 F capacitor and a 6350 resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current Io in the circuit? Calculate the circuit's time constant 7. How much time t must elapse from the closing of the circuit for the current to decrease to 3.39% of its initial value? Io = T = t = 2.36 x107 2.839 X10-2 3.45 Incorrect A SThe figure below shows a capacitor, with capacitance C = 35.0 μF, and a resistor, with resistance R = 40.0 kn, connected in series to a battery, with = 13.0 V. The circuit has a switch, which is initially open. + R (a) What is the circuit's time constant (in s)? (b) After the switch is closed for one time constant, how much charge (in C) is on the capacitor?A defibrillator is designed to pass a large current through a patient’s torso in order to stop dangerous heart rhythms. Its key part is a capacitor that is charged to a high voltage. The patient’s torso plays the role of a resistor in an RC circuit. When a switch is closed, the capacitor discharges through the patient’s torso. A jolt from a defibrillator is intended to be intense and rapid; the maximum current is very large, so the capacitor discharges quickly. This rapid pulse depolarizes the heart, stopping all electrical activity. This allows the heart’s internal nerve circuitry to reestablish a healthy rhythm.A typical defibrillator has a 32 μF capacitor charged to 5000 V. The electrodes connected to the patient are coated with a conducting gel that reduces the resistance of the skin to where the effective resistance of the patient’s torso is 100 Ω. If a patient receives a series of jolts, the resistance of the torso may increase. How does such a change affect the initial current…
- a. Write your understanding about the time constant for RC circuit. A capacitor is discharging through a resistor. If it takes a time T for the charge on a capacitor to drop to half its initial value, how long (in terms of T) does it take for the stored energy to drop to half its initial value? -e b. A 10-m-long wire has a resistance equal to 0.20 Q and carries a current equal to 5.0 A. (a) What is the potential difference across the entire length of the wire? (b) What is the electric-field strength in the wire?Consider the circuit shown below. The resistor has R= 40 ohms and the capacitor has C= 6 x10^-6 F. Initially the switch is open and there is no charge on the capacitor and no current in the resistor. At a time after the switch is closed the current in the resistor is 2 A and the charge on the capacitor is q= +3.00 x 10^-4 C. What is the emf of the battery?A capacitor that is initially uncharged is connected in series with a resistor and a 400.0 V emf source with negligible internal resistance. Just after the circuit is completed, the current through the resistor is 0.800 mA and the time constant for the circuit is 6.00 s. ▼ Part A What is the resistance of the resistor? Express your answer with the appropriate units. R = Submit Part B C = Value Submit What is the capacitance of the capacitor? Express your answer with the appropriate units. Request Answer 0 Value Units Request Answer Units ? ?
- The emf in the circuit shown in the figure below is 23.5 V. The resistors have resistances R1 = 46.6 0, R2 = 94.3 Q, and R3 = 53.0 N. R1 R3 (a) What is the current in the emf device? A (b) If a fourth resistor is added in parallel with the first three resistors, will the current in the emf device increase, decrease, or stay the same? O increase O decrease O stay the same (c) If, instead, a fourth resistor is added in series with the emf device, will the current in the emf device increase, decrease, or stay the same? O increase O decrease O stay the samedv + 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) =answer a and b only..a12