Question 12.4: Consider the circuit of Figure 12.9 in which a non-ideal vol......
Consider the circuit of Figure 12.9 in which a non-ideal voltage source is connected to a variable load resistor with resistance R_L. The source voltage is V and its internal resistance is R_S. Calculate the value of R_L which results in the maximum power being transferred from the voltage source to the load resistor.
Let i be the current flowing in the circuit. Using Kirchhoff’s voltage law and Ohm’s law gives
V=i\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)Let P be the power developed in the load resistor. Then,
P=i^2 R_{\mathrm{L}}=\frac{V^2 R_{\mathrm{L}}}{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^2}Clearly P depends on the value of R_L. Differentiating w.r.t. R_L and using the quotient rule, we obtain
\begin{aligned} \frac{\mathrm{d} P}{\mathrm{~d} R_{\mathrm{L}}} & =V^2 \frac{1\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^2-R_{\mathrm{L}} 2\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)}{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^4} \\ & =V^2 \frac{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)-2 R_{\mathrm{L}}}{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^3} \\ & =V^2 \frac{R_{\mathrm{S}}-R_{\mathrm{L}}}{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^3} \end{aligned}Equating \frac{\mathrm{d} P}{\mathrm{~d} R_{\mathrm{L}}} to zero to obtain the turning point gives
V^2 \frac{R_{\mathrm{S}}-R_{\mathrm{L}}}{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^3}=0that is,
R_{\mathrm{L}}=R_{\mathrm{S}}So a turning point occurs when the load resistance equals the source resistance. We need to check if this is a maximum turning point, so
\begin{aligned} \frac{\mathrm{d} P}{\mathrm{~d} R_{\mathrm{L}}} & =V^2 \frac{R_{\mathrm{S}}-R_{\mathrm{L}}}{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^3} \\ \frac{\mathrm{d}^2 P}{\mathrm{~d} R_{\mathrm{L}}^2} & =V^2 \frac{-1\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^3-\left(R_{\mathrm{S}}-R_{\mathrm{L}}\right) 3\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^2}{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^6} \\ & =V^2 \frac{-\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)-3\left(R_{\mathrm{S}}-R_{\mathrm{L}}\right)}{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^4} \\ & =V^2 \frac{2 R_{\mathrm{L}}-4 R_{\mathrm{S}}}{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^4} \\ & =2 V^2 \frac{\left(R_{\mathrm{L}}-2 R_{\mathrm{S}}\right)}{\left(R_{\mathrm{S}}+R_{\mathrm{L}}\right)^4} \end{aligned}When R_L = R_S, this expression is negative and so the turning point is a maximum. Therefore, maximum power transfer occurs when the load resistance equals the source resistance.