Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is

Answer 1

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is  $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

   The power rating of the bulb is $P = 100 \ W$

   The resistance is   $R = 143 \ \Omega$

   The  voltage is  $V = V_o sin [2 \pi ft]$

   The  energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

   The  voltage  $V_o = 110 \ V$

   The frequency is  $f = 60 \ Hz$

    The  time considered is  $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

             $P = \frac{V^2}{ R}$

=>          $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=>           $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So  

     $U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=>  $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=>  $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=>  $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=>  $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=>  $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=>   $U = 7.563 *10^{5} \ J$