Answer
Two bulbs can be connected.
Work Step by Step
We know that the equivalent resistance of the entire circuit is given as
$ R_{eq}=\frac{V}{I_{max}}$
The resistance of one light bulb is
$ R=\frac{V^2}{P}$
Let $ n $ be the number of light bulbs required for the circuit
$\implies \frac{1}{R_{eq}}=\frac{n}{R}$
This can be rearranged as:
$ n=\frac{R}{R_{eq}}$
$\implies n=(\frac{V^2}{P})(\frac{I_{max}}{V})$
$\implies n=\frac{VI_{max}}{P}$
We plug in the known values to obtain:
$ n=\frac{(85V)(2.1A)}{65W}$
$ n=2.7$
Thus, only two bulbs can be connected in parallel before the total current exceeds $2.1A $.
