Answer
The voltage, in mV, across the $1.0\Omega$ resistor is the value that is set by the switches.
Work Step by Step
We see that the $1.0\Omega$ resistor does not appreciably affect the equivalent resistance of the circuit, since it is thousands of times smaller in value.
Case 1, binary number 0001. The n = 0 switch is closed. The equivalent resistance of the circuit is $16.0k\Omega$. The current that flows is $I=\frac{16V}{16.0k\Omega }=1.0mA$.
The voltage across the $1.0\Omega$ resistor is $V = IR = (1.0mA)(1.0\Omega)=1.0mV$.
Case 2, binary number 0010. The n = 1 switch is closed. The equivalent resistance of the circuit is $8.0k\Omega$. The current that flows is $I=\frac{16V}{8.0k\Omega }=2.0mA$.
The voltage across the $1.0\Omega$ resistor is $V = IR = (2.0mA)(1.0\Omega)=2.0mV$.
Case 3, binary number 0100. The n = 2 switch is closed. The equivalent resistance of the circuit is $4.0k\Omega$. The current that flows is $I=\frac{16V}{4.0k\Omega }=4.0mA$.
The voltage across the $1.0\Omega$ resistor is $V = IR = (4.0mA)(1.0\Omega)=4.0mV$.
Case 4, binary number 1001. The n = 3 and the n = 0 switches are both closed. Find the equivalent resistance of the circuit.
$$R_{eq}= (\frac{1}{16.0k\Omega }+\frac{1}{2.0k\Omega })^{-1}=1.778k\Omega $$
The current that flows is $I=\frac{16V}{1.778k\Omega }=9.0mA$.
The voltage across the $1.0\Omega$ resistor is $V = IR = (9.0mA)(1.0\Omega)=9.0mV$.
We see that the analog voltage, in mV, across the $1.0\Omega$ resistor is the digital value set by the switches. This is a 4-bit digital-to-analog converter.
