New solution

Alexphysics edited
View solution History
revision #18868 newer →
en/11.1.30.md +46 −0
@@ -0,0 +1,46 @@
+### Statement
+
+$11.1.30.$ [Insert the problem statement]
+
+### Solution
+
+First, we analyze the total power supplied to the rotor (50 W), which is distributed as:
+
+· Friction losses: 4% of 50 W
+ P_{\text{friction}} = 0.04 \times 50 = 2\ \text{W}
+· Mechanical power converted into electrical power:
+ P_{\text{electrical}} = 50\ \text{W} - 2\ \text{W} = 48\ \text{W}
+
+Calculation of the EMF
+
+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:
+
+P_{\text{electrical}} = \frac{\mathcal{E}^2}{R}
+
+Solving for \mathcal{E}:
+
+\mathcal{E} = \sqrt{P_{\text{electrical}} \cdot R} = \sqrt{48 \times 300} = \sqrt{14400} = 120\ \text{V}
+
+\boxed{\mathcal{E} = 120\ \text{V}}
+
+Finding the new resistance while keeping the same speed
+
+The EMF depends only on the rotation speed (and the magnetic field). Since the rotation frequency remains the same, the EMF does not change:
+
+\mathcal{E}' = \mathcal{E} = 120\ \text{V}
+
+New resistance: R' = 60\ \Omega.
+New generated electrical power:
+
+P'_{\text{electrical}} = \frac{\mathcal{E}^2}{R'} = \frac{14400}{60} = 240\ \text{W}
+
+Friction losses depend on speed, which does not change, so they remain at 2 W.
+
+P_{\text{input}}' = P'_{\text{electrical}} + P_{\text{friction}} = 240\ \text{W} + 2\ \text{W} = 242\ \text{W}
+
+\boxed{P_{\text{input}}' = 242\ \text{W}}
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]