Question

1 henry =

• A. weber $\div$ ampere
• B. weber $\div$ volt
• C. weber ampere
• D. weber volt

Answer 1

Answer: (A)

weber $\div$ ampere

Answer 2

$E = - L \frac{dI}{dt}$ . Therefore $[L] = [E \,dt/dI]$ = volt second/ampere = volt second/(coulomb/second) = volt second$^2$/coulomb = joule second$^2$ / coulomb$^2$