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

The energy dissipated by a resistor is $10 \, mJ$ in $1 \, s$ when an electric current of $2 \, mA$ flows through it. The resistance is $....... \, \Omega$ (Round off to the Nearest Integer).

How many calories of heat will be produced approximately in a $210\, W$ electric bulb in $5$ minutes?

An electric heater of resistance $6 \, \Omega$ is run for $10 \, \text{minutes}$ on a $120 \, \text{V}$ line. The energy liberated in this period of time is:

Two bulbs are rated $(40 \; W, 200 \; V)$ and $(100 \; W, 200 \; V)$. What is the correct relation for their resistances?

An electric kettle takes $4\, A$ current at $220\, V$. How much time will it take to boil $1\, kg$ of water from room temperature $20\, ^\circ C$? The temperature of boiling water is $100\, ^\circ C$. (Assume specific heat of water $c = 4200\, J/kg\, ^\circ C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module