A series RL circuit includes a 905V battery a resistance of

A series RL circuit includes a 9.05-V battery, a resistance of R = 0.755 , and an inductance of L = 2.97 H. What is the induced emf 1.33 s after the circuit has been closed?

So, basically, I tried solving this equation using an RL circuit formula: I = emf/R(1-e^-t/(L/R)) and used V=IR to solve for the current, for which I plugged it back in to obtain the emf. I got an answer of about 31, which was wrong. If you could help me on this, it would make my day!

tau = L/R = 2.97/0.755 = 3.9337

V = V0*e^(-t/tau)

V = 9.05*e^(-1.33/3.9337)

V = 6.453 V

A series RL circuit includes a 9.05-V battery, a resistance of R = 0.755 , and an inductance of L = 2.97 H. What is the induced emf 1.33 s after the circuit has

A series RL circuit includes a 905V battery a resistance of

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