Answer
a) The voltage through $R_1$ will increase
The voltage through $R_2$ will increase from 0
The voltage through $R_3$ will decrease
The voltage through $R_4$ will decrease
b) The current through $R_1$ will increase
The current through $R_2$ will increase from 0
The current through $R_3$ will decrease
The current through $R_4$ will decrease
c) The power output will increase
d) Our predictions are correct.
Work Step by Step
a) When we close the circuit, we are adding a resistor that is in parallel, meaning the total resistance will decrease. Therefore, the current through the circuit will increase.
The voltage through $R_1$ will increase
The voltage through $R_2$ will increase from 0
The voltage through $R_3$ will decrease
The voltage through $R_4$ will decrease, because $R_2$, $R_3$, and $R_4$ are connected in parallel and the current needs to divide into 3 paths.
b) From Ohm's Law $V=IR$, we know current and voltage are proportional. If voltage increases, current increases.
The current through $R_1$ will increase
The current through $R_2$ will increase from 0
The current through $R_3$ will decrease
The current through $R_4$ will decrease
c) Power can be calculated by the following equations: $P=I^2R=\frac{V^2}{R}$ The current, voltage, and power output are proportional and the resistance is inversely proportional to power output. The total resistance decreases, so power output will increase.
d) Switch open
$R_{eq}=R_1+\frac{1}{\frac{1}{R_3}+\frac{1}{R_4}}=232.5\Omega$
$I_{1}=\frac{V}{R_{eq}}=0.0946A$
$V_3=V_4$, so $I_3=I_4$
$V_3=22.0V-(0.0946A)(155\Omega)=7.3V$
$I_3=\frac{V_3}{R_3}=\frac{7.3V}{155\Omega}=0.0473A$
$I_{1}=0.0946A$
$I_{2}=0A$
$I_3=I_4=0.0473A$
Switch closed
$R_{eq}=R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}}=207\Omega$
$I_{1}=\frac{V}{R_{eq}}=0.106A$
$V_2=V_3=V_4$, so $I_2=I_3=I_4$
$V_3=22.0V-(0.106A)(155\Omega)=5.5V$
$I_3=\frac{V_3}{R_3}=\frac{5.5V}{155\Omega}=0.0355A$
$I_{1}=0.106A$
$I_2=I_3=I_4=0.0355A$
Our predictions are correct.