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InterviewSolution
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A battery has an emf E and internal resistance r. A variable resistance R is connected across the terminals of the battery. Find the value of R such that (a) the current in the circuit id maximum (b) the potential difference across the terminals is maximum. |
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Answer» Current in the circuit, `I = E/(R + r)` Power deliverd to the resistance R is, `P = I^(2) xx R = (E^(2)R)/(R +r)^(2)` Power delivered to the load will be maximum when `R/(R +r)^(2)` is maximum, i.e., `(d)/(dR) [R (R + r)^(-2)]=0` or `[(R +r)^(-2) + R(-2) (R + r)^(-3)]=0` or `(R + r)^(-3) [r +R]=0` As `(R + r)^(-3)!= 0` for finite value of R, So, r-R =0 or R=r When R = r, the power transferred to the load is `= E^(2)r/(r+r)^(2) = E_(2)/4r = max`. |
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