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`100 J` of heat are produced each second in a `4 Omega` resistance. Find the potential difference across the resistor. |
Answer» `H=100J, R=4 Omega, t=1 s, V=?` From Eq. (12.21) we have the current through the resistor as `I= sqrt(H//Rt)` `=sqrt[100J//(4Omegaxx1s)]` =5A Thus the potential difference across the resistor, V [ from Eq. [(12.5)] is V=iR `=5 Axx4 Omega` = 20 V. |
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