Question

Determine the equivalent resistance of the "ladder" of equal 30 Ω resistors shown in figure, in other words, what resistance would an ohmmeter read if connected between points A and B? b) (5 points) What is the power dissipated in the upper resistors on the left (the one connected to point A) if a 41 V battery is connected between points A and B? 

Answer 1

a) Equivalent resistance is R' = R +R+ R = 3R R" - R RI (BR) - ZK RARI Rt3R R"= RR+R" = R +R+ 3R = UR RW er Blu Rant 402.) RIN. HR Req = R+R+RW = R+R HR Reg. 468 - 41630) = 822 kcal Current in circuit I - Vno . 49 15 15 6) 1. Ra I-15 V = 0.5 A 302 Pa I?R = 0.5A)? (302) (P = 7.5 W 1 (the one connected to point A)

Answer 2

Answer 3

SOLUTION :

a.

Starting from the right end loop : 

Reqv1 = (R*3R)/(R + 3R) = 3/4 R

Now, the middle loop becomes the end loop :

Reqv2 = (R * (2R+3/4 R) / (R + (2R + 3/4 R)

= 11/15 R 

Now, we have one loop only 

R eqv = (2R + 11/15 R) = 41/15 R 

When R = 30 Ω :

Ohm meter will read resistance of the circuit :

= R env 

= 41/15 R

= 41/15 * 30 

= 82 Ω (ANSWER).

b.

41 V applied between point A an B.

So,

Current, I,  at the upper entry point A

= V / R eqv

= 41 / 82

= 0.5 amp

So, power dissipated in the upper resistor R connected to point A

= I^2 * R 

= 0.5^2 * 30

= 7.5 W (ANSWER).