Here, $\mathcal{E}$ is the emf of the source and $R$ is the resistance of the resistor. We can rearrange this equation to solve for $R$:
\[R = \frac{\mathcal{E}}{I}\]
where $I$ is the current. Substituting the given values $\mathcal{E} = 110 \, \text{V}$ and $I =
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