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Two capacitors, having equal capacitance C, are having same charges q on each plate. They are connected across a resistor R as shown in the figure. What is the total heat generated in the resistor upto the instant when the steady state has reached |
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Answer» `q^(2)//2C` `q_(1)=CV, q_(2) = CV` where v is the potential difference between the plates. Also from charge conservation `q_(1) +q_(2) = 0` implies` V = 0` Thus, final energy stored in the capacitors`(1)/(2)CV^(2)+(1)/(2)CV^(2)=0` Initial energy stored =`(q^(2))/(2C)+q^(2)/(2C)=(q^(2))/(C)` Thus, heat developed in the RESISTOR = `q^(2)//C` |
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