how many 65w light - bulbs can be connected in parallel across a potential difference of 85v before the…

Explanation:

Step1: Find resistance of one light - bulb

Using $P = \frac{V^{2}}{R}$, where $P = 65W$ and assume the bulb is rated at its normal operating voltage (we can use $P$ and $V$ to find resistance). Rearranging for $R$, we get $R=\frac{V^{2}}{P}$. For a 65 - W bulb, if we assume it is rated for a common voltage (let's use the voltage across it in the circuit for power - resistance relationship), $R=\frac{V^{2}}{P}=\frac{85^{2}}{65}\Omega$.

Step2: Find the equivalent resistance of the parallel combination

Using Ohm's law $V = IR$, and we know $V = 85V$ and $I = 2.1A$. The equivalent resistance of the parallel combination $R_{eq}=\frac{V}{I}=\frac{85}{2.1}\Omega$.

Step3: Use the formula for parallel resistors

The formula for the equivalent resistance of $n$ identical resistors $R$ in parallel is $R_{eq}=\frac{R}{n}$. We know $R=\frac{85^{2}}{65}$ and $R_{eq}=\frac{85}{2.1}$. Substituting into $R_{eq}=\frac{R}{n}$, we get $n=\frac{R}{R_{eq}}=\frac{\frac{85^{2}}{65}}{\frac{85}{2.1}}$. Simplifying $n=\frac{85\times2.1}{65}\approx 2.7$.

Answer:

c. 2.7 lamps