Question

​How much energy is dissipated as heat during a two-minute time interval by a 1.5-k resistor which has a constant 20-V potential difference across its leads?

Answer 1

Energy dissipated = i^2 *R*t

i = 20/1500 = 1/75 A

So, Energy dissipated = i^2 *R*t

= (1/75)^2 * 1500*120 = 32J

Explanation:

Answer 2

Given:

A 1.5kΩ resistor is connected across a 20V source.

To Find:

The amount of heat dissipated as heat during the interval of 2 minutes.

Solution:

Heat dissipated in time $t$ by a resistance with a voltage across it $V$ and a current $I$ flowing through it,

      $H=VIt$   (1)

Using Ohm's law, $I=\frac{V}{R}$

Substituting the value of $I$ in equation (1) we get,

   $H=V(\frac{V}{R})t$

⇒ $H=\frac{V^{2}t }{R}$

⇒ $H=\frac{(20)^2*2*60}{1500}$

⇒ $H=32J$

Hence, heat dissipated across the resistor is 32J.