In the circuit of Fig. E26.15, each resistor represents a...

In the circuit of $\textbf{Fig. E26.15}$, each resistor represents a light bulb. Let $R_1 = R_2 = R_3 = R_4$ = 4.50 $\Omega$ and $\varepsilon$ = 9.00 V. (a) Find the current in each bulb. (b) Find the power dissipated in each bulb. Which bulb or bulbs glow the brightest? (c) Bulb $R_4$ is now removed from the circuit, leaving a break in the wire at its position. Now what is the current in each of the remaining bulbs $R_1$, $R_2$, and $R_3$ ? (d) With bulb $R_4$ removed, what is the power dissipated in each of the remaining bulbs? (e) Which light bulb(s) glow brighter as a result of removing $R_4$ ? Which bulb(s) glow less brightly? Discuss why there are different effects on different bulbs.

Transcript

00:01 Here in this given problem these four resistors are representing the four bulbs, four lamps in a mixed grouping along with a cell, a battery providing mfe.

00:26 This is the resistance r1, r2, r3 and r4 and all these resistors are having same value.

00:37 R1 is equal to r2 is equal to r3 is equal to r4 and all these are 4 .50 ohm.

00:48 The battery voltage this is 9 .00 volt.

00:54 In the first part of the problem we have to find current passing through all these four lamps.

01:01 So, first of all, these three r2, r3 and r4 are in parallel.

01:05 So, as r2, r3 and r4 are in parallel, three resistors are in parallel and all three are identical.

01:22 So, their net combination in parallel combination, their value rp will be given as either of the three means r2 and divided by the number and that is 3.

01:33 This is 4 .50 ohm divided by 3 means it comes out to be equal to 1 .50 ohm.

01:42 Now r1 that is in series with this parallel combination rp.

01:54 So equivalent resistance of this combination that will be given by r1 plus rp means this is 4 .50 plus 1 .50 ohm and it is calculated to be equal to 6 .00 ohm.

02:15 So, current passing through this combination that will be using ohms law will be given by epsilon by r.

02:22 R equivalent means this is 9 divided by 6 or we can say total current entering into the circuit.

02:29 This is 1 .5 ampere.

02:33 Now as current remains the same in series, so current passing through r1 means i1 that will be equal to total circuit current means 1 .5 ampere and as it is divided in parallel combination and all the three parallel resistors are identical.

03:01 So, we can say i2 will be equal to i3 will be equal to i4 and that is 1 .5 divided by 3.

03:11 So, all these three currents are calculated to be equal to 0 .5 ampere.

03:20 So, these are the answers for the first part of the given problem here.

03:29 Now, in the second part of the problem, we have to find power dissipated as heat along across these bulbs.

03:40 So, starting with p1 that is i1 square r1 means square of 1 .5 multiplied by 4 .5.

03:49 So, this p1 is calculated to be equal to 10 .125 watt and across other three bulbs as these are in parallel and identical...

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