Question

Determine the equivalent resistance of the "ladder" of equal 155-? resistors. In other words, what resistance would an ohmmeter read if connected between points A and B?
What is the current through resistor a if a 50.0-V battery is connected between points A and B?
What is the current through resistor b?
What is the current through resistor c

Answer 1

starting from right side

R,R and R are in series

R+R+R=3R

3R and R are in parallel

3R*R/(3R+R) =0.75R

R ,0.75R and R are in series

R+0.75R+R =2.75R

2.75R and R are in parallel

2.75R*R/(2.75R+R) =0.7333R

R,0.7333R and R are in series ,so equivalent resistance is

R_AB=R+0.7333R+R =2.7333R =2.7333*155

R_AB=423.67 ohms

b)

Current

I=V/R =50/423.67 =0.118 A

c)

Current through resistor b is

I_B=I*(2.75R/2.75R+R) =0.118*0.7333

I_B=0.08653 A

d)

I_C=0.118 A

Answer 2

SOLUTION :

a.

Let time of travel be t hr and distance covered be d km.

Rest time = 22 min = 22/60 =  0.3667 hr.

So, 

d = 99.5 * (t - 0.3667) 

Also, d = 77 * t

So,

99.5 (t - 0.3667) = 77 t

=> 99.5 t - 36.4866 = 77 t

=> 99.5 t - 77 t = 36.4866

=> 22.5 t = 36.4866

=> t = 36.4866 / 22.5 = 1.6216 hr = 1 hr 37 min 18 sec approx. (ANSWER).

b.

d = 77 * 1.6216 = 124.86 km approx. (ANSWER).

SOLUTION :

a.

Starting from the right end loop : 

R eqv1 = (R*3R)/(R + 3R) = 3/4 R

Now, the middle loop becomes the end loop :

R eqv2 = (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 

As  R = 155 Ω :

Ohm meter will read resistance of the circuit :

= R eqv. 

= 41/15 R

= 41/15 *  115

= 423.67Ω (ANSWER).

50 V applied between point A an B.

So,

b.

Current, Ia   through resistor a I.e. 155 Ω 

= V / R eqv

= 50 / 423.67

= 0.118 amp (ANSWER).

c.

Current, Ib   through resistor b  

= (V - Ia *2R) / R

= (50 - 0.118 *2*155)  / 155

=  0.0866 amp (ANSWER)

d.

Current, Ic   through resistor c I.e. 155 Ω 

= V / R eqv

= 50 / 423.67

= 0.118 amp (ANSWER).