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 [tex]U = \int\limits^T_0 {P(t)} \, dt[/tex]
Compute U if the bulb remains on for 5h
Answer:
The value is [tex]U = 7.563 *10^{5} \ J[/tex]
Explanation:
From the question we are told that
The power rating of the bulb is [tex]P = 100 \ W[/tex]
The resistance is [tex]R = 143 \ \Omega[/tex]
The voltage is [tex]V = V_o sin [2 \pi ft][/tex]
The energy expanded is [tex]U = \int\limits^T_0 {P(t)} \, dt[/tex]
The voltage [tex]V_o = 110 \ V[/tex]
The frequency is [tex]f = 60 \ Hz[/tex]
The time considered is [tex]t = 5 \ h = 18000 \ s[/tex]
Generally power is mathematically represented as
[tex]P = \frac{V^2}{ R}[/tex]
=> [tex]P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}[/tex]
=> [tex]P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}[/tex]
So
[tex]U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt[/tex]
=> [tex]U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt[/tex]
=> [tex]U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt[/tex]
=> [tex]U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt[/tex]
=> [tex]U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.[/tex]
=> [tex]U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.[/tex]
[tex]U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ][/tex]
=> [tex]U = 7.563 *10^{5} \ J[/tex]
