Question

challenge for brainly​

Answer 1

Answer:-

T = 120°C

Given :-

$R_a = 3 \Omega \\ R_b = 4\Omega$

$\alpha_a = 6 \times 10^{-3}K^{-1}\\ \alpha_b = 2 \times 10^{-3}K^{-1}$

To find :-

At what temperature the resistance of wires will be equal.

Solution:-

Let the temperature at which resistance will equal be T.

The Relation of temperature with resistance is given by :-

$\huge \boxed{R = R_{ref}[1+ \alpha (T-T_{ref})]}$

where,

• R is resistance
• T is temperature
• $T_{ref}$ temperature at reference 20°C.
• $\alpha$ Temperature coefficient.

→$R_a= 3 [ 1 + 6 \times 10^{-3} (T - 20)]$

→$R_a= 3[ 1 + 0.006 (T-20) ]$

→$R _a= 3 [ 1+0.006 (T-20) ]$

→$R_a = 3 (1+ 0.006T - 0.12)$

→$R_a = 3( 0.006T +0.88)$

→$R_a = 0.018T +2.64$

Now,

→$R_b = 4[1 + 2 \times 10^{-3} (T-20) ]$

→$R_b = 4 [ 1 + 0.002 (T-20) ]$

→$R_b = 4(1 + 0.002T - 0.04)$

→$R_b = 4( 0.96 +0.002T)$

→$R_b = 3.84 + 0.008T$

• since, at that temperature resistance is equal.

→$R_a = R_b$

→$0.018T + 2.64 = 3.84 + 0.008T$

→$0.018 T -0.008T = 3.84 -2.64$

→$0.01T = 1.2$

→$T = \dfrac{1.2}{0.01}$

→$T = 120^{\circ}C$

hence,

The resistance is equal at 120°C