18. (II) (a) Determine the equivalent resistance of the "ladder" of equal 175-? resistors shown

Name: 18. (II) (a) Determine the equivalent resistance of the "ladder" of equal 175-? resistors shown in Fig. 19–49. In other words, what resistance would an ohmmeter read if connected between points A and B? (b) What is the current through each of the three resistors on the left if a 50.0-V battery is connected between points A and B?
Uploaded: 2022-05-15T16:49:50-08:00
Duration: 5 min 59 s
Channel: Suman K
Description: 18. (II) (a) Determine the equivalent resistance of the "ladder" of equal 175-? resistors shown in Fig. 19–49. In other words, what resistance would an ohmmeter read if connected between points A and B? (b) What is the current through each of the three resistors on the left if a 50.0-V battery is connected between points A and B?

Given the nature of the problem, it seems we're dealing with an infinite ladder of resistors, ea

18. (II) (a) Determine the equivalent resistance of the...

18. (II) (a) Determine the equivalent resistance of the "ladder" of equal 175-? resistors shown in Fig. 19–49. In other words, what resistance would an ohmmeter read if connected between points A and B? (b) What is the current through each of the three resistors on the left if a 50.0-V battery is connected between points A and B?

Transcript

00:01 Hello students in this question we have to determine the equivalent resistance of the ladder which is given and the value of each resistance is 175 oms and in the b part we have to find the current through each of the three resistors on the left if a 50 -volt battery is connected between points a and b so let's see the diagram first so here this is a ladder so in the a part we have three resistors in series here so when three resistance are in series so they add up so r plus r plus r is 3 r now this r1 will be parallel with the this this r so this will be suppose r 2 so r2 will be r1 multiplied by r over r1 plus r so here the value of r1 is 3 r multiplied by r multiple by r will be 3 r square over this will be 3 r plus r so this will be 3 r plus r so this will be 3 over 3 over 4 r so this will be 3 over 4 r so r2 will be now r2 will be in series with this r and this r.

01:08 So that is suppose r3.

01:10 So it will be r plus r2 plus r.

01:15 So it is r plus 3 over 4r plus r.

01:21 So if we take 4 lcm, so 4r plus 3r plus 4r plus 4r.

01:27 So this is 4 plus 4 8 plus 3 is 11 over 4.

01:32 R this is r3 this is 11 over 4 r now this 11 over 4 r is in parallel with this r r 4 will be equal to r3 multiplied by r over r3 plus r so r3 is 11 over 4 r square over this is 11 over 4 r plus r so here we have 11 over 4 r square over we have 4 lcm so 11 r plus 4 r so this is 11 r square over 4 multiplied 4 over 15 r so 4 and 4 are cancelled so we have 11 over 15 r so now this r 4 is in parallel with this r and this r so are effective will be r plus r 4 plus r so it will be here r plus 11 over 15 r plus r so we take 15 and then so this is 15 r plus 11 r plus 15 15 is 30 plus 11 is 41 over 15 r this is the effective resistance now we put the value of r so are effective is equal to 41 over 15 so, r effective will be equal to, so here r comes out to be 477 .75 oms.

03:16 Now in the b part, it has been asked that the current through the three resistors on the left when a battery of 50 volt is connected across a and b.

03:26 So we have to find the current through these three registers.

03:30 So first of all, we find the total current that is passed, that is coming out of the battery.

03:35 So we know that here v is equal to i r.

03:40 So current i will be v over r.

03:43 So the total current will be 50 over the effective resistance that is 477 .75.

03:51 So here the value of i comes out to be 0 .105 amperes...