R_1 and R_2 are two resistors made of differe | vedclass

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

Question

(D) Let the resistances at $0^{\circ}C$ be $R_{01}$ and $R_{02}$.At temperature $	heta$,the resistances are given by:$R_1 = R_{01}(1 + \alpha 	heta)

Vedclass · Accepted Answer

$\beta/\alpha$

Vedclass · Answer

(D) Let the resistances at $0^{\circ}C$ be $R_{01}$ and $R_{02}$.At temperature $	heta$,the resistances are given by:$R_1 = R_{01}(1 + \alpha 	heta)$$R_2 = R_{02}(1 - \beta 	heta)$The total resistance in series is $R_s = R_1 + R_2$.$R_s = R_{01}(1 + \alpha 	heta) + R_{02}(1 - \beta 	heta)$$R_s = (R_{01} + R_{02}) + 	heta(R_{01}\alpha - R_{02}\beta)$For the total resistance $R_s$ to be independent of temperature $	heta$,the coefficient of $	heta$ must be zero:$R_{01}\alpha - R_{02}\beta = 0$$R_{01}\alpha = R_{02}\beta$Therefore,the ratio of the resistances is $\frac{R_{01}}{R_{02}} = \frac{\beta}{\alpha}$.

Similar Questions

In the given circuit,the potential difference between points $P$ and $Q$ is . . . . . . (in $\text{ V}$)

The charge flowing through a resistance $R$ varies with time $t$ as $Q = at - bt^2$,where $a$ and $b$ are positive constants. The total heat produced in $R$ is

The equivalent resistance of the following infinite network of resistances is

Find the current in the $3\,\Omega$ resistance in the given circuit.

$A$ galvanometer of resistance $G$ is connected in a circuit. Now,a resistance $R$ is connected in series with the galvanometer. To keep the main current in the circuit unchanged,the resistance $S$ to be put in parallel with the series combination of $G$ and $R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module