Introduction:
When bulbs are connected in series, the total resistance increases and the current passing through each bulb is the same. On the other hand, when bulbs are connected in parallel, the total resistance decreases and the voltage across each bulb is the same.

Given information:
- Three identical bulbs are connected in series and dissipate a power p.
- When the bulbs are connected in parallel, the power dissipated is np.

Analysis:
In a series circuit, the total resistance (R) is the sum of the individual resistances (R1, R2, R3, ...):
R = R1 + R2 + R3

The power dissipated by a bulb is given by:
P = V^2 / R

Case 1: Bulbs in series
When the bulbs are connected in series, the total resistance (R) increases and the current passing through each bulb is the same. Therefore, the power dissipated by each bulb is the same.

Let's assume the power dissipated by each bulb in series is P1.

Since the bulbs are identical, the power dissipated by each bulb is the same:
P1 = P / 3

Case 2: Bulbs in parallel
When the bulbs are connected in parallel, the total resistance (R) decreases and the voltage across each bulb is the same. Therefore, the power dissipated by each bulb is different.

Let's assume the power dissipated by each bulb in parallel is P2.

Since the voltage across each bulb is the same, and power is given by P = V^2 / R, we can write:
P2 = (V^2 / R) = (V^2 / (R / 3)) = 3V^2 / R

Substituting the value of R from the series circuit equation:
P2 = 3V^2 / (R1 + R2 + R3)

Since the bulbs are identical, R1 = R2 = R3 = R, we can write:
P2 = 3V^2 / (3R) = V^2 / R

Comparing the power dissipated in parallel (P2) and series (P1) circuits:
P2 = P1 / 3

Since the power dissipated in parallel is given as np, we have:
np = P1 / 3

Therefore, the value of n is 1/3.

N=9