Question

The small transformers that power many consumer products produce a$12.0V_{rms}$, $60Hz$emf. Design a circuit using resistors and capacitors that uses the transformer voltage as an input and produces a $6.0V_{rms}$ output that leads the input voltage by $45°$.

Short answer

The$R_1=6\Omega ,R_2=2.5\Omega andC=3.12\times {10}^{-4}F$.

Step-by-step solution

Step 1:Given Information 

We are given that small transformer produces voltage $12V_{rms},60Hz$emf.

We have to design the circuit which have $6V_{rms}$output that leads the input voltage by$45°$.

Step 2:Formula Used

As the capacitor voltage is lead by the resistor voltage . So, the input must ne in between the resistors and the phase difference is of $45°$.

Now, $Z_R=Z_C$

then $V_C=V_R=\frac{12}{\sqrt{2}}$

which gives$V_C=V_R=8.5V$.

Step 3:Simlify

Let $R_1$be the resistor needed to make the voltage $6V_{rms}$

then $\frac{R_1}{6}=\frac{Z_R}{8.5}$

On simplifying we get,

$R_1=0.71Z_R$

let role="math" localid="1650018615730" $Z_R=8.5\Omega$then

On solving, $R_1=6\Omega and{R}_2=2.5\Omega$

and $Z_C=8.5\Omega =\frac{1}{\omega C}$

where $\omega =2\mathrm{\pi }\times 60$

On solving we get,

$C=3.12\times {10}^{-3}F$

Hence, the resistor$R_1=6\Omega and{R}_2=2.5\Omega$and the capacitor$C=3.12\times {10}^{-3}F$.