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
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
RAB=R+0.7333R+R =2.7333R =2.7333*155
RAB=423.67 ohms
b)
Current
I=V/R =50/423.67 =0.118 A
c)
Current through resistor b is
IB=I*(2.75R/2.75R+R) =0.118*0.7333
IB=0.08653 A
d)
IC=0.118 A
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).
Determine the equivalent resistance of the "ladder" of equal 155-? resistors. In other words, what resistance...
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Tulsiram Garg Wed, Dec 8, 2021 8:13 AM