Q. 265.0( 1 Vote )

Consider the situation shown in figure. The frame is made of the same material and has a uniform cross-sectional are everywhere. Calculate the amount of heat flowing per second through a cross-section of the bent part if the total heat taken out per second from the end at 100°C is 130J.

Answer :

For the above frame, redrawing the fig,



The equivalent resistance network becomes –

Let RAB, RBC, RCD, RDE, REF and RBE be the equivalent resistance across each cross-section as shown in fig. below –



Resistance in terms of conductivity

R= (1)

From fig (1) and (2)and equation (1)

RAB = , RBC = , RCD =, RDE =, REF= and RBE =

Now, lets reduce the network into equivalent network .

Since RBC RCD RDE are connected in series, let R1 be their equivalent resistance.

Then R1 = RBC+RCD + RDE



Now the circuit reduces to –


Now from Kirchhoff’s current law(KCL), we know The algebraic sum of all currents entering and exiting a node must equal zero

Hence ,KCL at point E , since -

q = q1 +q2

Now, since R1 and RBE are in parallel, so total heat across R1 and RBE will be same.

ie, q1R1 = q2RBE


q2 =

Now, q = q1 +q2

= q1 + q1

Given q = 130 J, substituting above

130 =

q1 = 60 J

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