Q. 20 C4.5( 12 Votes )

# Determine the equivalent resistance of networks shown in Fig. 3.31.

(a)

It can be observed from the given circuit that in the first small loop, two resistors

of resistance 1 Ω each are connected in series.

Hence, their equivalent resistance = (1 + 1) = 2 Ω

It can also be observed that two resistors of resistance 2 Ω each are connected in series.

Hence, their equivalent resistance = (2 + 2) = 4 Ω.

Now the given circuit diagram can be explained as

Now draw 4 loops each having 2 resistors of 2 Ω and 4 Ω connected in parallel.

Hence Equivalent Resistance Re of each loop is

Re = 4/3 Ω

Since all 4 loops will be connected in series and each loop will have 4/3 Ω resistance

Hence, equivalent resistance of the given circuit is 4 × 4/3 = 16/3 Ω .

(b)

From the given circuit that five resistors of resistance R each are connected in series.

Hence, equivalent resistance of the circuit = R + R + R + R + R = 5R

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Ohm's Law - Full Concept in one Go59 mins
Know about Internal Resistance & EMF42 mins
Interactive Quiz on Ohm's law & Drift + Thermal Velocity42 mins
Interactive Quiz on Current Electricity46 mins
An Elaborative lecture on basics of Current Electricity53 mins
Quick Revision of Current Electricity68 mins
Quiz | Wheatstone Bridge51 mins
Quiz | Drift Velocity & Thermal Velocity33 mins
Drift Velocity & Thermal Velocity34 mins
Kirchoff's Law for VoltageFREE Class
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses