document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=907585>Question 907585</A>: <I><SPAN STYLE=\"word-break: break-all;\"> for a resistor in a direct circuit that does not vary its resistance, the power that a resistor must dissipate is directly proportional tp the square of the voltage across the resistor. the resistor must dissipate 1/16 watt power when the voltage across the resistor is 14 volts. find the power the resistor must dissipate when the voltage across it is 42 volts </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #550690 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=550690\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> For a resistor in a direct circuit that does not vary its resistance, the power that a resistor must dissipate is directly proportional to the square of the voltage across the resistor.<BR>\n" );
document.write( "The resistor must dissipate 1/16 watt power when the voltage across the resistor is 14 volts.<BR>\n" );
document.write( "Find the power the resistor must dissipate when the voltage across it is 42 volts.<BR>\n" );
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document.write( "The power equation for this problem<BR>\n" );
document.write( "P = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=E%5E2%2FR ALIGN=MIDDLE  ALT=\"E%5E2%2FR\"   DISABLED_style= \"max-width:90%; height:auto;\" > where E = the voltage, R = resistance in ohms, P in watts<BR>\n" );
document.write( "Find the resistance<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F16 ALIGN=MIDDLE  ALT=\"1%2F16\"   DISABLED_style= \"max-width:90%; height:auto;\" > = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=14%5E2%2FR ALIGN=MIDDLE  ALT=\"14%5E2%2FR\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F16 ALIGN=MIDDLE  ALT=\"1%2F16\"   DISABLED_style= \"max-width:90%; height:auto;\" >R  = 196<BR>\n" );
document.write( "multiply both sides by 16<BR>\n" );
document.write( "R = 3136 ohms<BR>\n" );
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document.write( "When voltage is 42 volts<BR>\n" );
document.write( "P = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=42%5E2%2F3136 ALIGN=MIDDLE  ALT=\"42%5E2%2F3136\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "P = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1764%2F33136 ALIGN=MIDDLE  ALT=\"1764%2F33136\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "P = .5625 watts (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=9%2F16 ALIGN=MIDDLE  ALT=\"9%2F16\"   DISABLED_style= \"max-width:90%; height:auto;\" >) Watts<BR>\n" );
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document.write( "We can note here that when you triple the voltage, you increase the power by 3^2\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );