A 10 ohm resistor and 10 henry inductor are attached in series to a 10 volt DC supply. If the current i(0)2.5 amperes, what is the current after 1 second? Round in the tenths place.I got 1.55 but that isn't right.
Strictly speaking, you should write the differential equation and solve it for the initial condition, i(0) 2.5, but there's an easier way: The steady state current will eventually decrease to just V/R 10V/10Ω 1 amp, which is less that the initial value, so the current will decay exponentially from 2.5 to 1 amp, a difference of 1.5 amps: i (t) 1 + 1.5e^(-Rt/L) After 1 second: i(1) 1 + 1.5e^(-10/10) 1 + 1.5e^-1 1.55 amps But you forgot to round to the tenths place: i(1) 1.6 amps Unless your instructor is a real jerk and voltage is in the opposite direction? So now the current must fall exponentially from 2.5 amps to -1 amp: i (t) -1 + 3.5e^(-Rt/L) After 1 second: i(1) -1 + 3.5e^(-10/10) -1 + 3.5e^-1 0.29 amps Round to the tenths place: i(1) 0.3 amps
Question doesn't make sense. Current after a long time would be 10/10 1 amp, where just the resistance would be relevant. Current when the supply is first turned on will be less, not more than 1 amp.
What do you want to know?
