phy electric circuits問題

Let n be the no. of light bulbs and R be the resistance of each bulb.

In the parallel connection, currents through each bulb are equal (because they are identical). Thus current through each bulb = 2/n A

Applying Ohm's Law to a bulb
15 = (2/n).R
i.e. R = 15n/2

In the series connection, the total resistance of n bulbs = nR
Using Ohm's Law again, 15 = (5/1000).nR
i.e. 15 = (5/1000)n(15n/2)
n^2 = 2 x 1000/5 = 400
n = 20

1. unknown number of identical light bulb are connected to a 15V in parallel, The current through the battery is 2A
so. R = V / I : 15V / 2A = 7.5Ω

2. If the light bulbs are connected to the battery in series, the current through the battery is 5mA
so. R = V / I : 15V / 0.005A = 3000Ω

3. 3000Ω / 7.5Ω = 400, √400 = 20bulb

4. if 3000Ω is 20 bulb connected in series
so. 3000Ω / 20 = 150Ω each bulb

5. if 7.5Ω is 20 bulb connected in parallel
so. 7.5 x 20 = 150Ω each bulb

battery: 15V, 2A
bulb: 15V, 5mA

In parallel circuit
It=I1+I2...In

so the no. of bulb= current of battery/ current of bulb
= 2A/5mA
=2/0.005
=400
this circuit connects to 400 bulbs in parallel.