There is some flow from the positive terminal, which enters the network that con

Question

There is some flow from the positive terminal, which enters the network that consists of bulbs 1, 2, and 3. Some flow will leave that network too, and enters the network that consists of bulbs 4 and 5.

a) Suppose that the flow out of the network of bulbs 4 and 5 is of size 1 glow. How large is the flow through bulb 4 and through bulb 5? Explain

b) If the flow out of the network of bulbs 4 and 5 is of size 1 glow, what is the flow into and out of the network of bulbs A, C, and D? Explain

c) Given the flow into and out of the network of bulbs 1, 2, and 3 in the previous question, compare the sizes of the flow through bulb 1, through bulb 2, and through bulb 3. Then compare the sizes of thess flows to the size of the flow through bulb 4.

d) Using the previous answers, rank the bulbs in order of decreasing brightness. Put and 'equal' sign between bulbs of the same brightness.

Explanation / Answer

Let the voltage given out by the battery (VT) = 5 V

Let the resistances of each bulb = 2 (since they are all identical)

We calculate the total resistance (RT) first.

Considering the bulbs 2 and 3, they are in series.

So, R23 = 2 + 2 = 4

Now, the combination of bulbs 2 and 3; and bulb 1 are in parallel.

So, 1/R123 = 1/R1 + 1/R23

= 1/2 + 1/4

=> 1/R123 = 3/4

So, R123 = 4/3

Now the bulbs 4 and 5 are in parallel.

So, 1/R45 = 1/R4 + 1/R5

= 1/2 + 1/2

So, R45 = 1 (as reciprocal of 1 is also 1).

Finally we have 2 resistances R123 and R45 . They are in Series.

So, RT = R123 + R45

= 4/3 + 1

=> RT = 7/3

We now calculate the total current in the circuit (IT).

By Ohm's Law, we have:

V = I * R

Put values,

IT = 5 / (7/3)

=> IT = 15/7 A.

Now we find the voltages and currents across each bulbs.

We will follow this rule :

Series: The same amount of current flows through each component. The voltage drop on each component is determined by the resistance (or reactance for AC) of the component.


Parallel: The same amount of voltage is applied to each component, and the current flowing through a component is determined by its resistance (or reactance).

Note :

1st Box refers to the circuit having bulbs 1,2 and 3.

2nd Box refers to the circuit having bulbs 4 and 5.

The 1st and the 2nd boxes are in series.

Hence I1st Box = 15/7 A and I2nd Box = 15/7 A.

For 1st Box,

R123 = 4/3

So, by Ohms Law,

V1st Box = (15/7) * (4/3) = 20/7 V.

VT = 5 V.

So, V2nd Box = VT -  V1st Box

= 5 - (20/7)

=> V2nd Box = 15/7 V.

Consider the 1st Box :

As the combination of bulbs 2 and 3; and bulb 1 are in parallel.

V1 = 20/7 V.

V23 = 20/7 V.

For Bulb 1:

Using Ohm's Law,

V1 = I1 * R1

Put values and solve,

I1 = (20/7) / 2 = 10/7 A.

For combination of Bulbs 2 and 3 :

V23 = 20/7 V.

R23 = 4

By Ohm's Law,

I23 = (20/7) / 4 = 5/7 A.

As bulbs 2 and 3 are in series,

I2 = I3 = I23 = 5/7 A.

V2 = I2 * R2

= (5/7) * 2

=> V2 = 10/7 V.

V2 + V3 = V23

=> V3 = (20/7) - (10/7) = 10/7 V.

Consider the 2nd Box :

V2nd Box = 15/7 V.

As bulbs 4 and 5 are in parallel.

V4 = V5 = V2nd Box = 15/7 V.

For Bulb 4:

By Ohm's Law,

I4 = V4 / R4

Put values,

I4 = (15/7) / 2 = 15/14 A.

As bulbs 4 and 5 are in parallel,

I5 = 15/14 A.

Now we have the required data.

Flow here implies Current.

Glow here implies Power given out by bulb.

a) Flow or Current through Bulb 4 (I4) = Flow or Current through Bulb 5 (I5)= 15/14 A.

b) Flow or Current into 1st Box = Flow or Current out of 1st Box = 15/7 A.

c)

Flow or Current through Bulb 1 (I1) = 10/7 A.

Flow or Current through Bulb 2 (I2) = 5/7 A.

Flow or Current through Bulb 3 (I3) = 5/7 A.

Flow or Current through Bulb 4 (I4) = 15/14 A.

Clearly,

I1 > I4 > I2 = I3

d) For brightness, we calculate Power.

As, Power (P) = Voltage(V) * Current(I)

P1 = V1 * I1 = (20/7) * (10/7) = 200/49 W = 4.08 W.

P2 = V2 * I2 = (10/7) * (5/7) = 50/49 W = 1.02 W.

P3 = V1 * I3 = (10/7) * (5/7) = 50/49 W = 1.02 W.

P4 = V4 * I4 = (15/7) * (15/14) = 225/98 W = 2.29 W.

P5 = V5 * I5 = (15/7) * (15/14) = 225/98 W = 2.29 W.

Clearly,

P1 > P4 = P5 > P2 = P3

Hence, P1 is the brightest of all while P2 and P3 are the dullest of all.

Hope this helps.

All The Best!!!

Please do rate me if the answer is correct. :)

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