Solution

To solve this 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 $2 W$.

$P_{\text{input}}' = P'_{\text{electrical}} + P_{\text{friction}} = 240\ \text{W} + 2\ \text{W} = 242\ \text{W}$

$\boxed{P_{\text{input}}' = 242\ \text{W}}$