Which of the four resistances will generate the greatest amount of heat when current flows from X to Y in the network shown here(see image)? a)P= 2ohm b)Q= 4ohm c)R= 1ohm d)S= 2ohm please give reason.

In series system ,resistance with larger value generates more heat compared to lower resistance,because current is same in both resistance and heat = i² R.

In parallel system resistance with lower value genrates more heat compared to higher resistance,because potential across both resistance is same and heat=V²/R.

So, In the given problem

more heat will be generated in lower branch with R & S ;compared to upper branch with P & Q ; because (R+S)<(P+Q) : (parallel system).

In lower branch more heat will be generated in S compared to R , since S>R (series system)

 

Therefore S will generate greatest amount of heat.

 

Better Method:

Let current through point X (or Y) is i.

   Resistance between X and Y  R =

therefore potential difference between X and Y is

V_xy = 20*i

Let current through branch P & Q is i₁ and through branch R & S is i₂ .

then

i₁ *(P+Q)= V_xy

             = 20*i  

and

i₂ *(R+S)=V_xy

             = 20*i

on solving we get

 i₁ = i/3

 i₂ =2i/3     (Note that i₁ +i₂ = i : total current)

 

Now heat generated by S and Q is

 

Clearly   H_S > H_Q