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When a battery is connected to the resistance of 10 Omega the current in the circuit is 0.12 A . The same battery gives 0.07A current with 20 Omega Calculate e.m.f and internal resistance of the battery. |
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Answer» SOLUTION :We know that E= IR + IR `I_1 r+ I_1 R_1 = I_2 r + I_2 R_2` `impliesI_1 r - R_2 r = I_2 R_2 - I_1 R_1` `r(I_1 - I_2) = I_2 R_2 - I_1 R_1` `impliesr= (I_2 R_2 - I_1 R_1)/(I_1 - I_2)` `r= (0.07xx20- 0.12xx10)/(0.12-0.07)` `=(1.4-12)/(0.05)` `=(0.2)/(0.05) = 4 Omega` INTERNAL resistance `r= 4 Omega` e.m.f `E= I_1 r_1 + I_1 R_1` `0.12xx4 + 0.12xx10= 0.48+1.2` `E= 1.68 ` volt |
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