Physics Solution Q30

Solution to Question 30

Step 1: Analyze Current and Power Relation

The power rating of a bulb is given by $P_{\text{rated}} = \frac{V^2}{R_b}$, where $V$ is the mains voltage and $R_b$ is the bulb resistance. Assuming the resistance of the connecting wires $r$ is negligible compared to the bulb resistance ($r \ll R_b$), the current in the circuit is primarily determined by the bulb. $$ I \approx \frac{P_{\text{rated}}}{V} $$

Step 2: Relate Wire Power Dissipation to Bulb Rating

The power dissipated in the connecting wires is given by $P_{\text{wire}} = I^2 r$. Substituting the expression for current: $$ P_{\text{wire}} = \left( \frac{P_{\text{rated}}}{V} \right)^2 r = \left( \frac{r}{V^2} \right) P_{\text{rated}}^2 $$ Since $V$ and $r$ are constant, the power dissipated in the wires is directly proportional to the square of the rated power of the bulb: $$ P_{\text{wire}} \propto P_{\text{rated}}^2 $$

Step 3: Calculate for the New Bulb

Given:
$P_{\text{rated},1} = 50 \text{ W}$ leads to $P_{\text{wire},1} = 1.0 \text{ mW}$.
$P_{\text{rated},2} = 100 \text{ W}$.

Using the proportionality: $$ \frac{P_{\text{wire},2}}{P_{\text{wire},1}} = \left( \frac{P_{\text{rated},2}}{P_{\text{rated},1}} \right)^2 $$ $$ \frac{P_{\text{wire},2}}{1.0 \text{ mW}} = \left( \frac{100 \text{ W}}{50 \text{ W}} \right)^2 = (2)^2 = 4 $$ $$ P_{\text{wire},2} = 4 \times 1.0 \text{ mW} = 4.0 \text{ mW} $$

Final Answer

The power dissipated in the connecting wires with the 100 W bulb will be 4.0 mW.