Solution Found!
The input to the circuit shown in Figure DP 3-5 is the voltage source voltage vs. The
Chapter 3, Problem DP 3-5(choose chapter or problem)
The input to the circuit shown in Figure DP 3-5 is the voltage source voltage \(v_{\mathrm{s}}\). The output is the voltage \(v_{\mathrm{o}}\). The output is related to the input by
\(v_{\mathrm{o}}=\frac{R_2}{R_1+R_2} v_{\mathrm{s}}=g v_{\mathrm{s}}\)
The output of the voltage divider is proportional to the input. The constant of proportionality, g, is called the gain of the voltage divider and is given by
\(g=\frac{R_2}{R_1+R_2}\)
The power supplied by the voltage source is
\(p=v_{\mathrm{s}} i_{\mathrm{s}}=v_{\mathrm{s}}\left(\frac{v_{\mathrm{s}}}{R_1+R_2}\right)=\frac{v_{\mathrm{s}}^2}{R_1+R_2}=\frac{v_{\mathrm{s}}^2}{R_{\mathrm{in}}}\)
where
\(R_{\text {in }}=R_1+R_2\)
is called the input resistance of the voltage divider.
(a) Design a voltage divider to have a gain, g = 0.65.
(b) Design a voltage divider to have a gain, g = 0.65, and an input resistance, \(R_{\text {in }}=2500 \Omega\).
Questions & Answers
QUESTION:
The input to the circuit shown in Figure DP 3-5 is the voltage source voltage \(v_{\mathrm{s}}\). The output is the voltage \(v_{\mathrm{o}}\). The output is related to the input by
\(v_{\mathrm{o}}=\frac{R_2}{R_1+R_2} v_{\mathrm{s}}=g v_{\mathrm{s}}\)
The output of the voltage divider is proportional to the input. The constant of proportionality, g, is called the gain of the voltage divider and is given by
\(g=\frac{R_2}{R_1+R_2}\)
The power supplied by the voltage source is
\(p=v_{\mathrm{s}} i_{\mathrm{s}}=v_{\mathrm{s}}\left(\frac{v_{\mathrm{s}}}{R_1+R_2}\right)=\frac{v_{\mathrm{s}}^2}{R_1+R_2}=\frac{v_{\mathrm{s}}^2}{R_{\mathrm{in}}}\)
where
\(R_{\text {in }}=R_1+R_2\)
is called the input resistance of the voltage divider.
(a) Design a voltage divider to have a gain, g = 0.65.
(b) Design a voltage divider to have a gain, g = 0.65, and an input resistance, \(R_{\text {in }}=2500 \Omega\).
ANSWER:
Step 1 of 4
a)
Given ,
