A battery of 24 V and internal resistance \( 2\,\Omega \) is connected to a load (external resistance) \( R \). The value of current so that power delivered to the load will be maximum, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6 A
The power delivered to the load is
\( \displaystyle P = I^{2}R = \frac{E^{2}R}{(R + r)^{2}} \)
Setting \( dP/dR = 0 \) gives \( (R + r)^{2} - 2R(R + r) = 0 \), that is \( R = r \). So the power in the load is maximum when the load resistance equals the internal resistance (maximum power transfer).
\( R = r = 2\,\Omega \)
\( E = 24\,\mathrm{V} \)
\( \displaystyle I = \frac{E}{R + r} = \frac{24}{2 + 2} = 6\,\mathrm{A} \)