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JEE Advanced 1998 · Units and Measurements

The SI unit of inductance, the henry can be written as

Answers: A. weber/ampere\text{weber}/\text{ampere} · B. volt-sec/amp\text{volt-sec}/\text{amp} · C. Joule/(ampere)2\text{Joule}/(\text{ampere})^{2} · D. ohm-second\text{ohm-second}

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Written solution

Step 1: Use the magnetic flux relation.

The magnetic flux Φ\Phi is proportional to the current II.

Φ=LI\Phi = L I L=ΦIL = \frac{\Phi}{I} [L]=weberampere[L] = \frac{\text{weber}}{\text{ampere}}

Step 2: Use the induced electromotive force relation.

The induced electromotive force ee is related to the rate of change of current.

e=−LdIdte = -L \frac{dI}{dt} L=∣e∣∣dIdt∣L = \frac{|e|}{\left|\frac{dI}{dt}\right|} [L]=volt⋅secondampere[L] = \frac{\text{volt} \cdot \text{second}}{\text{ampere}}

Step 3: Use the stored magnetic energy relation.

The energy UU stored in an inductor depends on the current.

U=12LI2U = \frac{1}{2} L I^{2} L=2UI2L = \frac{2U}{I^{2}} [L]=jouleampere2[L] = \frac{\text{joule}}{\text{ampere}^{2}}

Step 4: Use the RLRL circuit time constant relation.

The time constant τ\tau of a series RLRL circuit depends on resistance RR and inductance LL.

τ=LR\tau = \frac{L}{R} L=RτL = R \tau [L]=ohm⋅second[L] = \text{ohm} \cdot \text{second}

Step 5: Compare the units.

All four derived units represent the henry (H\text{H}).

1 H=weberampere=volt⋅secondampere=jouleampere2=ohm⋅second1 \text{ H} = \frac{\text{weber}}{\text{ampere}} = \frac{\text{volt} \cdot \text{second}}{\text{ampere}} = \frac{\text{joule}}{\text{ampere}^{2}} = \text{ohm} \cdot \text{second}

Answer: The henry (H\text{H}) can be written as weberampere\dfrac{\text{weber}}{\text{ampere}}, volt⋅secondampere\dfrac{\text{volt} \cdot \text{second}}{\text{ampere}}, jouleampere2\dfrac{\text{joule}}{\text{ampere}^{2}}, or ohm⋅second\text{ohm} \cdot \text{second}.

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