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| Course | Level | Date | Clip# | Line | Interlocutor | Content |
|---|---|---|---|---|---|---|
| Oral Assessments | Graduate | 2019-08-22 | 4 | 235 | S01 | so we get back to the (.) definition of- (.) ∆final definition |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 236 | S01 | of ohms law∆↘ |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 237 | UNK | (0.7) |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 238 | S01 | ⁇which is⁇ v equals i r→ |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 239 | UNK | (1.7) |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 240 | S01 | and as the current is not zero→ |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 241 | UNK | (0.6) |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 242 | S01 | you see that uh the current is not zero→ so we can divide |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 243 | UNK | (1.0) |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 244 | S01 | so r is v by i↘ |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 245 | UNK | (1.3) |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 246 | S01 | so: (5.6) which is: (0.7) ten ohms↘ |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 247 | UNK | (2.6) |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 248 | S01 | so→ we had a bulb (0.5) we know the resistance of the bulb→ |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 249 | UNK | (0.6) |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 250 | S01 | xxx is ten ohms |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 251 | UNK | (0.9) |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 252 | S01 | so ∆if at all∆ if (0.6) u:h (.) we know the resistance→ |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 253 | UNK | (0.5) |
| Oral Assessments | Graduate | 2019-08-22 | 4 | 254 | S01 | and ∆if at all∆ if we connect to higher supply |