A battery is connected in series with a 3.0 ohm resistor and a 12 mH inductor, the max current in the circuit is 150 mA. What is the time constant of this circuit, and what is the EMF of the battery?If the time constant (T) is equal to the time required to reach 63.2 % of the max current, I calculated it to be the time required to reach 0.0948 Amps. T (0.632)(0.00150 A) 0.0948 A. But this is as far as I got. The equation I found that I think I should use is I E/R (1-e ^(-t/T) Please help - thank you
The battery is effectively related to an 18 ohm resistor. VIR, 36I(18) So the present out of the battery is two amps. Now evaluate the 12 ohm resistor. PI^2R (2)^2*1240 8 W. you're wonderful if the subject is looking with regard to the flexibility dissipated via the 12 ohm resistor. The 6 ohm resistor could expend 24W. books could be incorrect, purely seem on the Bible : ) If the two resistors have been related to the battery in parallel, their valuable resistance could be 4 ohms. PV^2/R 36^2/6 216W. you're suitable. PI(I)R might want for use for this subject too, yet you will possibly desire to calculate I in the process the 6 ohm resistor. the present by using each resistor is diverse.) PVV/R might want for use for the 1st subject, yet you will possibly desire to calculate V in the time of purely the 12 ohm resistor first.
The time constant is simply L/R. That's a standard result (Like RC is the time constant for a resistor/capacitor). Time constant 12x10^-3H / 3.0ohm 4 x10^-3s 4ms When the current is steady, the inductor - if it's perfect - behaves like a short-circuit - i.e. zero resistance. This is because the magnetic field inside it isn't changing so there is no induced emf. The toal circuit resistance is 3.0 ohms and the current is 150mA. So cell's emf IR 150x10^-3A x 3.0ohm 450x10^-3V 0.45V
