Question

Problem IIIA A pair of resistors Ri and R2 share the wattage of W watts totally dissipated by them in the circuit of Figure. III-A a. Determine the currents ll and 12 in terms of Ri, R2, and W. (Ans: 1,-(WRP)12 × [R2/(RI+R2)] A ) Hint: Use current division theorem b. Suppose R1-30 and R2-30, denote the values of R, and R respectively at room temperature 30°C. These elements are made of two different conducting materials of conductivities, Ơi and Ơ2 siemen/m (respectively) at room temperature (30°C); and, they are of identical geometry (of length L meter and cross sectional area A m2). The temperature coefficients of resistivity of the heating elements, R, and R2 are α/°C and α2 /C respectively Determine the values of the conductance, G, and G2 at 100°C. Ans: G1-100-G1-30/(1 +70%), unit ?; G2-100-7] 6 c. In Fig. 3, the current (I) through R is set to control a current-controlled voltage- operated switch (trip-switch); and V,-β11. Suppose the switch/trip operates when ν- VThreshold. Determine the corresponding threshold temperature (TThreshold in terms of V, Vyreshold, β, σι, α, and A/L), at which the switch will trip FIGURE 111-A 12

answer (B) only

show all work

Answer 1

Dear student the as requested, the answer to part B can be obtained as

We know that resistance $R=l /\sigma A$.Here l and A are same for both resistors but consuctivity is different, so we have two resistors of 2 different resistance values

Further we know resistance of a resistor varies with temperature. With increase in temperature, the vibration of the atoms increases resulting in increased rate of collisions. This increased rate of collisions hinders the movement of carriers resulting n increase in resistance of the resistor.

It has been found that resistance increases with temperature as

$R_{T} = R_{o}[1 + \alpha \Delta t]$

$R_{T}$ is resistance at certain temperature T (here 100°C)

  $R_{o}$ is resistance at some known temperature (Here is 30°C)

  $\alpha$ is temperature coefficient

$\Delta t$ is the difference in temperature

Here Ro is $R_{1-30}$ and  $\Delta t$ =100 -30 = 70°C

Temperature coefficients are given for both resistors

So R1 =  $R_{1-30}$[1 + $\alpha_{1}$70] and

R2 = $R_{2-30}$[1 + $\alpha _{2}$ 70]

Now we need to calculate conductance (G) which is reciprocal of resistance(R).

Hence G1= 1/ R1 = $1 / R_{1-30}[1+70 \alpha _{1}]$

$G_{1-100}= G_{1-30}/(1+70 \alpha _{1})\Omega ^{-1}$ [ ]

Also

G2 =1/ R2 $=1/ R_{1-30}(1+70 \alpha _{2})$

$G_{2-100}=G_{2-30}/(1+70 \alpha _{2})\Omega ^{-1}$ [

G2-30 1/R2-30