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| ### Statement |
| ### Statement |
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| $11.1.30.$ [Insert the problem statement] |
| $11.1.30.$ |
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| What kind of EMF does a DC dynamo develop if, with a circuit resistance of |
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| 300 ohms, 50 watts of power is used to rotate the rotor, and the friction loss is |
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| 4% of the power? How much power is needed to maintain the same frequency |
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| with a circuit resistance of 60 ohms? |
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| ### Solution |
| ### Solution |
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| To solve this first, we analyze the total power supplied to the rotor (50 W), which is distributed as: |
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| First, we analyze the total power supplied to the rotor (50 W), which is distributed as: |
| · Friction losses: $4% of 50 W$ |
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| $P_{\text{friction}} = 0.04 \times 50 = 2\ \text{W}$ |
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| · Friction losses: 4% of 50 W |
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| P_{\text{friction}} = 0.04 \times 50 = 2\ \text{W} |
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| · Mechanical power converted into electrical power: |
| · Mechanical power converted into electrical power: |
| P_{\text{electrical}} = 50\ \text{W} - 2\ \text{W} = 48\ \text{W} |
| $P_{\text{electrical}} = 50\ \text{W} - 2\ \text{W} = 48\ \text{W}$ |
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| Calculation of the EMF |
| Calculation of the EMF |
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| Assuming the internal resistance of the dynamo is negligible (or that the 300 Ω already includes the entire circuit), all the generated electrical power is dissipated in the external resistance R: |
| Assuming the internal resistance of the dynamo is negligible (or that the 300 Ω already includes the entire circuit), all the generated electrical power is dissipated in the external resistance R: |
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| P_{\text{electrical}} = \frac{\mathcal{E}^2}{R} |
| $P_{\text{electrical}} = \frac{\mathcal{E}^2}{R}$ |
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| Solving for \mathcal{E}: |
| Solving for $\mathcal{E}$: |
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| \mathcal{E} = \sqrt{P_{\text{electrical}} \cdot R} = \sqrt{48 \times 300} = \sqrt{14400} = 120\ \text{V} |
| $\mathcal{E} = \sqrt{P_{\text{electrical}} \cdot R} = \sqrt{48 \times 300} = \sqrt{14400} = 120\ \text{V}$ |
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| \boxed{\mathcal{E} = 120\ \text{V}} |
| $\boxed{\mathcal{E} = 120\ \text{V}}$ |
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| Finding the new resistance while keeping the same speed |
| Finding the new resistance while keeping the same speed |
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| The EMF depends only on the rotation speed (and the magnetic field). Since the rotation frequency remains the same, the EMF does not change: |
| The EMF depends only on the rotation speed (and the magnetic field). Since the rotation frequency remains the same, the EMF does not change: |
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| \mathcal{E}' = \mathcal{E} = 120\ \text{V} |
| $\mathcal{E}' = \mathcal{E} = 120\ \text{V}$ |
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| New resistance: R' = 60\ \Omega. |
| New resistance:$ R' = 60\ \Omega$. |
| New generated electrical power: |
| New generated electrical power: |
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| P'_{\text{electrical}} = \frac{\mathcal{E}^2}{R'} = \frac{14400}{60} = 240\ \text{W} |
| $P'_{\text{electrical}} = \frac{\mathcal{E}^2}{R'} = \frac{14400}{60} = 240\ \text{W}$ |
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| Friction losses depend on speed, which does not change, so they remain at 2 W. |
| Friction losses depend on speed, which does not change, so they remain at $ 2 W$. |
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| P_{\text{input}}' = P'_{\text{electrical}} + P_{\text{friction}} = 240\ \text{W} + 2\ \text{W} = 242\ \text{W} |
| $P_{\text{input}}' = P'_{\text{electrical}} + P_{\text{friction}} = 240\ \text{W} + 2\ \text{W} = 242\ \text{W}$ |
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| \boxed{P_{\text{input}}' = 242\ \text{W}} |
| $\boxed{P_{\text{input}}' = 242\ \text{W}}$ |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| $\boxed{\mathcal{E} = 120\ \text{V}}$ |
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| $\boxed{P_{\text{input}}' = 242\ \text{W}}$ |
