1824 New Hampshire gubernatorial election
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The 1824 New Hampshire gubernatorial election was held on March 9, 1824, in order to elect the governor of New Hampshire. Former Democratic-Republican United States senator from New Hampshire David L. Morril defeated incumbent Democratic-Republican governor Levi Woodbury and former Federalist governor Jeremiah Smith. Since no candidate received a majority in the popular vote, Morril was elected by the New Hampshire General Court per the state constitution.[1]
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March 9, 1824
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 County results Morril: 40â50% 50â60% 60â70% Woodbury: 50â60% 60â70% | |||||||||||||||||||||
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General election
On election day, March 9, 1824, Democratic-Republican candidate David L. Morril won the popular vote by a margin of 3,244 votes against his foremost opponent and incumbent Democratic-Republican governor Levi Woodbury. But since no candidate received a majority of the popular vote, a separate election was held by the New Hampshire General Court, which chose Morril as the winner, thereby retaining Democratic-Republican control over the office of governor. Morril was sworn in as the 10th governor of New Hampshire on June 2, 1824.[2]
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic-Republican | David L. Morril | 14,985 | 49.19 | |
| Democratic-Republican | Levi Woodbury (incumbent) | 11,741 | 38.54 | |
| Federalist | Jeremiah Smith | 3,300 | 10.83 | |
| Scattering | 438 | 1.44 | ||
| Total votes | 30,464 | 100.00 | ||
| Democratic-Republican hold | ||||