Question

Consider a source of voltage $\mathit{v}\left(t\right)\mathbf{=}\mathbf{10}\sqrt{\mathbf{2}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{t}\mathbf{}\mathbf{}\mathit{V}$, with an internal resistance of $\mathbf{1800}\mathbf{}\mathit{\Omega }$. A transformer that can be considered as ideal is used to couple a $\mathbf{50}\mathbf{}\mathit{\Omega }$resistive load to the source. (a) Determine the transformer primary-to-secondary turns ratio required to ensure maximum power transfer by matching the load and source resistances. (b) Find the average power delivered to the load, assuming maximum power transfer.

Short answer

Answer

(a)The turns ratio is $a=6$.

(b)The average power delivered to the load is .

Step-by-step solution

Determine the formulas load resistance referred to primary and the average power delivered to the load.

Determine the turns ratio.

Consider that $R_L$is a load resistance connected across the secondary winding and $R_L\text{'}$is load resistance referred to the primary.

The turns ratio is given by $a=\frac{N_1}{N_2}$.

The relation of load resistance at secondary and referred to primary is,

${\mathit{R}}_{\mathbf{L}}\mathbf{\text{'}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}{\mathit{R}}_{\mathbf{L}}$ ……. (1)

The average power delivered to the load is given by,

${\mathit{P}}_{\mathbf{L}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{L}}^{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{L}}^{\mathbf{\text{'}}}}$ ……. (2)

The voltage across load using voltage divider rule is given by,

${\mathit{V}}_{\mathbf{L}}\mathbf{=}{\mathit{V}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\left(\frac{R_L^{\text{'}}}{R_L^{\text{'}}+R_s}\right)$ ……. (3)

Determine the turns ratio.

(a)

Determine the turns ratio.

For maximum power transfer to the load, the internal resistance $R_S$should be same as load resistance referred to primary that is $R_L^{\text{'}}$.

$$\begin{gathered} R_L^{\text{'}}=R_S \\ {a}^2{R}_L=R_S \end{gathered}$$

Substitute$1800\Omega$for $R_S$and $50\Omega$for$R_L$in above equation.

$$\begin{gathered} a^2\times 50=1800 \\ a=6 \end{gathered}$$

Therefore, the turns ratio is 6.

Determine the average power delivered to the load.

(b)

The average power is transferred to the load, assuming maximum power transfer, hence $R_L\text{'}=R_S=1800\Omega$.

By the voltage distribution method, the voltage across $R_L\text{'}$using equation (3) is,

role="math" localid="1655875806989" $$\begin{gathered} V_L=\left(\frac{10\sqrt{2}}{\sqrt{2}}\right)\times \frac{1800\Omega }{1800\Omega +1800\Omega } \\ =5V \end{gathered}$$

Determine the average power.

Substitute $5V$for $V_L$and $1800\Omega$for $R_L\text{'}$in equation (2).

$$\begin{gathered} P_L=\frac{5^2}{1800} \\ =13.9mW \end{gathered}$$