Solve q no 14 urgently

Solve q no 14 urgently (D) both me bubs fuse Q An external resistance R is connected to a cen E and .nternal power developed •n the external Circuit is nEX'rnwn wten 13 / An electric heater has a resistance Of 150 and can bear a maxinunn current of ../ the heater is to be used On 220 V mms. the least resistance requ•red in the circuit (A) 70 A cell Of E M F E and resistance r supplies currents for the same time t through external res'stances R, and Rz respectively It the heat produced •n both cases the same. then the internal resistance is

Dear Student,

internal resistance is r For first resistance R_{1}, Heat, H = I^{2} R_{1} t Since, I = \frac{E}{(R_{1} + r}) Thus, H = [\frac{E}{R_{1} + r}]^{2} R_{1} t - - - - (i) And for second resistance R_{2}, H' = I'^{2} R_{2} t H' = [\frac{E}{R_{2} + r}]^{2} R_{2} t - - - - (ii) According to question, H = H' Or [\frac{E}{R_{1} + r}]^{2} R_{1} t = [\frac{E}{R_{2} + r}]^{2} R_{2} t or \frac{R_{2}}{(R_{2} + r)^{2}} = \frac{R_{1}}{(R_{'} + r)^{2}} On solving we get, R = \sqrt{R_{1} R_{2}} r e g a r d s