Question:medium
Which option(s) represents/represent the dielectric loss tangent of a substrate?
Which option(s) represents/represent the dielectric loss tangent of a substrate?
Show Hint
Loss tangent compares the in-phase (lossy) current to the out-of-phase (reactive) current: \(\tan\delta=(\omega\epsilon''+\sigma)/(\omega\epsilon')\).
Updated On: Jul 20, 2026
- Ratio of the real to imaginary parts of the total displacement current
- \(\left(\omega\epsilon''+\sigma\right)/\left(\omega\epsilon'\right)\)
- Ratio of the electric susceptibility to permittivity
- Ratio of the polarization vector \(\vec{P}\) to the displacement vector \(\vec{D}\)
Show Solution
The Correct Option is A, B
Solution and Explanation
Step 1: Start from the constitutive relation with loss.
Write the permittivity as $\epsilon = \epsilon' - j\epsilon''$ and add a real conductivity $\sigma$ for ohmic loss. The full current density becomes $J = (\sigma + j\omega\epsilon)E$.
Step 2: Split into real and reactive pieces.
\[ J = \sigma E + j\omega\epsilon' E + \omega\epsilon'' E = (\sigma + \omega\epsilon'')E + j\omega\epsilon' E \]
The first bracket is in phase with $E$, this is where power is actually used up. The second is $90^\circ$ out of phase, this part just stores and returns energy.
Step 3: Define the loss tangent as this ratio.
By definition, loss tangent is in-phase current over out-of-phase current, or real over imaginary:
\[ \tan\delta = \frac{\sigma + \omega\epsilon''}{\omega\epsilon'} \]
This matches both statement (A), the ratio of real to imaginary parts of the total current, and statement (B), the algebraic form $(\omega\epsilon''+\sigma)/(\omega\epsilon')$.
Step 4: Test the remaining options.
Susceptibility to permittivity ($\chi/\epsilon$) and polarization to displacement ($|P|/|D|$) are both purely real ratios built from magnitudes at a single instant. Neither carries the $90^\circ$ phase lag that defines a loss, so neither is the loss tangent.
Step 5: Conclude.
\[ \boxed{\text{options (A) and (B)}} \]
Write the permittivity as $\epsilon = \epsilon' - j\epsilon''$ and add a real conductivity $\sigma$ for ohmic loss. The full current density becomes $J = (\sigma + j\omega\epsilon)E$.
Step 2: Split into real and reactive pieces.
\[ J = \sigma E + j\omega\epsilon' E + \omega\epsilon'' E = (\sigma + \omega\epsilon'')E + j\omega\epsilon' E \]
The first bracket is in phase with $E$, this is where power is actually used up. The second is $90^\circ$ out of phase, this part just stores and returns energy.
Step 3: Define the loss tangent as this ratio.
By definition, loss tangent is in-phase current over out-of-phase current, or real over imaginary:
\[ \tan\delta = \frac{\sigma + \omega\epsilon''}{\omega\epsilon'} \]
This matches both statement (A), the ratio of real to imaginary parts of the total current, and statement (B), the algebraic form $(\omega\epsilon''+\sigma)/(\omega\epsilon')$.
Step 4: Test the remaining options.
Susceptibility to permittivity ($\chi/\epsilon$) and polarization to displacement ($|P|/|D|$) are both purely real ratios built from magnitudes at a single instant. Neither carries the $90^\circ$ phase lag that defines a loss, so neither is the loss tangent.
Step 5: Conclude.
\[ \boxed{\text{options (A) and (B)}} \]
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