Two conductors have the same resistance at 0, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the resistance of both conductors at $0\,^{\circ}C$ be $R_0$. The resistance at temperature $\Delta t$ is given by $R = R_0(1 + \alpha \Delta

Vedclass · Accepted Answer

$\frac{\alpha_1 + \alpha_2}{2}, \frac{\alpha_1 + \alpha_2}{2}$

Vedclass · Answer

(A) Let the resistance of both conductors at $0\,^{\circ}C$ be $R_0$. The resistance at temperature $\Delta t$ is given by $R = R_0(1 + \alpha \Delta t)$.For the series combination: $R_s = R_1 + R_2 = R_0(1 + \alpha_1 \Delta t) + R_0(1 + \alpha_2 \Delta t) = 2R_0(1 + \frac{\alpha_1 + \alpha_2}{2} \Delta t)$. Comparing this with $R_s = 2R_0(1 + \alpha_s \Delta t)$,we get $\alpha_s = \frac{\alpha_1 + \alpha_2}{2}$.For the parallel combination: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_0(1 + \alpha_1 \Delta t)} + \frac{1}{R_0(1 + \alpha_2 \Delta t)}$. Using the binomial approximation $(1+x)^{-1} \approx 1-x$,we get $\frac{1}{R_p} \approx \frac{1}{R_0}(1 - \alpha_1 \Delta t + 1 - \alpha_2 \Delta t) = \frac{2}{R_0}(1 - \frac{\alpha_1 + \alpha_2}{2} \Delta t)$.Thus,$R_p \approx \frac{R_0}{2}(1 + \frac{\alpha_1 + \alpha_2}{2} \Delta t)$. Comparing this with $R_p = \frac{R_0}{2}(1 + \alpha_p \Delta t)$,we get $\alpha_p = \frac{\alpha_1 + \alpha_2}{2}$.

Two conductors have the same resistance at $0\,^{\circ}C$ but their temperature coefficients of resistance are $\alpha_1$ and $\alpha_2$. The respective temperature coefficients of their series and parallel combinations are nearly

Similar Questions

Some bulbs are connected in series to a $220\, V$ supply. If one bulb gets fused and the remaining bulbs are reconnected in series to the same supply,the brightness in the room will:

Six wires, each of resistance $r$, are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is

Eight resistances each of resistance $5\,\Omega$ are connected in the circuit as shown in the figure. The equivalent resistance between $A$ and $B$ is

Three resistances of equal values $R$ are arranged in four different configurations as shown in the figure. The power dissipation in increasing order is:

$A$ ring is made of a wire having a resistance $R_0 = 12\,\Omega$. Find the positions of points $A$ and $B$ such that the equivalent resistance of the circuit shown is $8/3\,\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module