Normal 0false false falseEN-US X-NONE X-NONE/* Style Definitions */table.MsoNormalTable{mso-style-name:Table Normal;mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-parent:;mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin-top:0in;mso-para-margin-right:0in;mso-para-margin-bottom:10.0pt;mso-para-margin-left:0in;line-height:115%;mso-pagination:widow-orphan;font-size:11.0pt;font-family:Calibri,sans-serif;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:Times New Roman;mso-bidi-theme-font:minor-bidi;}A 4 uF capacitor is charged to 400 volts and then its plates are joined through a resistance of 1 kW. The heat produced in the resistance is(a) 0.16 J(b) 1.28 J(b) 0.641(d) 0.32 J
Simran Bhatia , 11 Years ago
Grade 11
2 Answers
Hrishant Goswami
Last Activity: 11 Years ago
(d)
The energy stored in the capacitor
This energy will be converted into heat in the resistor.
Rishi Sharma
Last Activity: 4 Years ago
Dear Student, Please find below the solution to your problem.
The energy stored in the capacitor = 0.5CV^2 = 0.5*4*10^-4*(400)^2 = 0.32 J This energy will be converted into heat in the resistor.
Thanks and Regards
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