6 points there are two circuits below in one two light bulbs...

[6 Points] There are two circuits below. In one, two light bulbs are connected in series to a battery. In the other, one light bulb is connected to an identical battery. The three light bulbs are identical. (a) Which light bulb(s) give off the most light? . . . . . . . . . . . . . . . . . . . . (b) Which light bulb(s) give off the least light? . . . . . . . . . . . . . . . . . . . . . (c) Which light bulb(s) have the greatest current? . . . . . . . . . . . . . . . . .

Transcript

00:01 Hi, in the question we are given three circuits with identical batteries with voltage b and identical bulbs with resistances are on each of the bulb.

00:15 We are asked to rank the bulb according to the brightness.

00:20 So let us calculate the current flowing through each bulb because the brightness is depending on the current flowing through the bulb.

00:28 So circuit 1 we will get the current i is equal to v by r.

00:35 In second circuit we will get current i is equal to voltage by r equivalent we have two resistors connected in series so r equivalent is equal to 2r so the current through each bulb is equal to v by 2r.

00:56 In the third circuit we will get the current i is equal to again it is v by r equivalent.

01:03 The two resistors are connected in parallel therefore r equivalent is equal to r by 2 and we will get the current i is equal to v r by 2.

01:13 So it is 2v by r.

01:15 We can write the current through each bulb i a is equal to v by r i b is equal to i c.

01:27 Is equal to v by 2 r and i d is equal to i e is equal to 2 v by r by 2 that is v by r.

01:43 We are dividing this by 2 because the current is equally divided among these two branches because they are having same resistances.

01:54 Now let us compare the brightness.

01:56 Since a, d and e have same current flowing through it, the brightness of a, d and e will be same and in b and c we have less current flowing through it.

02:10 Therefore it will be less brighter than a, d and e.

02:15 Hence we can rank the bulbs as a is equal to d is equal to e then b is equal to c.

02:25 So based on brightness of the bulb bulbs a is equal equal to d is equal to e which is brighter than b equal to c.

02:33 Now coming to the second part of the question.

02:37 In the second part we are given a circuit with bulbs a, b, c and d and we are asked to analyze the bulbs and rank the bulb according to their brightness.

02:52 So let i be the current flowing through the circuit.

02:55 Then i current will flow through a.

02:57 It will get split into b and c if you will be the current...

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