A current i passes through a wire of length l | vedclass

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

Question

(A) The rate of heat generation is given by the electrical power $P = i^2 R$.We know that the resistance $R$ of a wire is given by $R = \rho \frac{l}{

Vedclass · Accepted Answer

$\frac{i^2 l \rho}{\pi r^2}$

Vedclass · Answer

(A) The rate of heat generation is given by the electrical power $P = i^2 R$.We know that the resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$, where $A$ is the cross-sectional area.The area of the cross-section of the wire is $A = \pi r^2$.Substituting the value of $A$ into the resistance formula, we get $R = \frac{\rho l}{\pi r^2}$.Now, substituting $R$ into the power formula, we get $P = i^2 \left( \frac{\rho l}{\pi r^2} \right) = \frac{i^2 l \rho}{\pi r^2}$.

$A$ current $i$ passes through a wire of length $l$, radius of cross-section $r$ and resistivity $\rho$. The rate of heat generation is

Similar Questions

$A$ lamp of $60 \, W, 230 \, V$ is used for $8 \, \text{hours}$. If the cost of $1 \, \text{unit}$ is $1.25 \, \text{Rs}$, what will be the total bill in $Rs$?

$A$ coil develops heat of $800 \, cal/sec$ when $20 \, V$ is applied across its ends. The resistance of the coil is .............. $\Omega$ $(1 \, cal = 4.2 \, J)$.

Two electric bulbs have power ratings $P_1$ and $P_2$ at voltage $V$. They are connected in parallel to a voltage source $V$. What is the total power consumed?

If voltage across a bulb rated $220 \text{ V} - 100 \text{ W}$ drops by $2.5\%$ of its rated value,the percentage of the rated value by which the power would decrease is ............... $\%$

An expression for the rate of heat generated,if a current of $I$ ampere flows through a resistance of $R$ $\Omega$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module