Begin by finding the resistance of the parallel connection:

\(\displaystyle \frac{1}{Z}=\frac{1}{Z_1+Z_2}+\frac{1}{Z_3}\)

\(\displaystyle \frac{1}{Z}=\frac{1}{32\Omega}+\frac{1}{16\Omega}\)

\(\displaystyle \frac{1}{Z}=\frac{3}{32\Omega}\)

\(\displaystyle Z=\frac{32}{3}\Omega\)

The total current is then found using Ohm's law:

\(\displaystyle I=\frac{V_{in}}{Z}\)

\(\displaystyle I=\frac{12V}{\frac{32}{3}\Omega}\)

\(\displaystyle I=1.125A\)

The symbol for a capacitor is written as a break in the circuit separated by two parallel lines of equal length as shown below. This loosely resembles the most common type of capacitor, a parallel plate capacitor.

First, we need to calculate the equivalent resistance of the three resistors in parallel. To do this, we will use the following equation:

\(\displaystyle \frac{1}{R_{eq}} = \sum\frac{1}{R} = \frac{1}{5}+\frac{1}{5}+\frac{1}{5} = \frac{3}{5}\)

\(\displaystyle R_{eq} = \frac{5}{3}\Omega\)

Now, to get the total equivalent resistance, we can simply add the two remaining values, since they are in series:

\(\displaystyle R_{total} = 3\Omega+\frac{5}{3}\Omega = \frac{14}{3}\Omega\)

First, we need to calculate the current flow through R2 without the extra voltage attached. We will need to calculate the total equivalent resistance of the circuit. Since the two resistors are in series, we can simply add them.

\(\displaystyle R_{eq} = R1 + R2 = 3\Omega+2\Omega=5\Omega\)

Then, we can use Ohm's law to calculate the current through the circuit:

\(\displaystyle V = IR\)

\(\displaystyle I=\frac{V}{R}=\frac{12V}{5\Omega} = 2.4A\)

Now that we have the current, we can calculate the additional current that the new voltage contributes:

\(\displaystyle I_{tot}=I+I_{new}\)

\(\displaystyle I_{new}=4A\)

\(\displaystyle I_{new} = 4A-2.4A= 1.6A\)

There is only one resistor (R2) in the path of the new voltage, so we can calculate what that voltage needs to be to deliver the new current:

\(\displaystyle V=IR = (1.6A)(2\Omega)=3.2V\)

First, let's remind ourselves that the effective resistance of resistors in a series is \(\displaystyle R_{eff}=R_{1}+R_{2}+...\) and the effective resistance of resistors in parallel is \(\displaystyle R_{eff}=\frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+...}\).

Start this problem by determining the effective resistance of resistors 2, 3, and 4:

\(\displaystyle R_{x}=R_{2}+R_{3}+R_{4}=3\Omega +2\Omega+1\Omega=6\Omega\) (This is because these three resistors are in series.)

Now, the circuit can be simplified to the following:

Next, we will need to determine the effective resistance of resistors \(\displaystyle x\) and 6:

\(\displaystyle R_{y}=\frac{1}{\frac{1}{R_{x}}+\frac{1}{R_{6}}}=\frac{1}{\frac{1}{6\Omega }+\frac{1}{2\Omega }}=1.5\Omega\)

Again, the circuit can be simplified:

From here, the effective resistance of the DC circuit can be determined by calculating the effective resistance of resistors \(\displaystyle y\), 1, and 5:

\(\displaystyle R_{eff}=R_{1}+R_{y}+R_{5}=1\Omega+1.5\Omega+4\Omega=6.5\Omega\)

The first step to figuring out this problem is to figure out how resistances of light bulb correlate to the power rating. For a resistor, the power dissipated is:

\(\displaystyle P=IV=\frac{V^{2}}{R}\)

Thus, there is an inverse relationship between the resistance of the lightbulb and the power rating. 

The second step is to take a look at circuit elements in series and in parallel. In series, they share the same current; in parallel they share the same voltage. Thus, for the two lightbulbs in series, the one with the higher resistance (lower wattage) will be brighter, and for a parallel configuration the one with the lower resistance (higher wattage) will be brighter.

The total resistance of resistors in a series is the sum of their individual resistances. In this case,

\(\displaystyle R_{tot}=R_1+R_2+...+R_n\)

\(\displaystyle R_{tot}=5\Omega +10\Omega + 15\Omega=30\Omega\)

Resistors in series:

\(\displaystyle R_s = R_1+R_2+R_3\)

\(\displaystyle R_s=3\Omega+3\Omega+3\Omega\)

\(\displaystyle R_s=9\Omega\)

Resistors in parallel:

\(\displaystyle \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

\(\displaystyle \frac{1}{R_p}=\frac{1}{3\Omega}+\frac{1}{3\Omega}+\frac{1}{3\Omega}\)

\(\displaystyle \frac{1}{R_p}=\frac{3}{3\Omega}\)

\(\displaystyle R_p=1\Omega\)

\(\displaystyle R_s-R_p=8\Omega\)

The equation for equivalent resistance for multiple resistors in parallel is:

\(\displaystyle \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}\)

Plug in known values and solve. 

\(\displaystyle \frac{1}{R_{eq}}=\frac{1}{100\Omega}+\frac{1}{10\Omega}+\frac{1}{1\Omega}\)

\(\displaystyle \frac{1}{R_{eq}}=\frac{111}{100\Omega}\)

\(\displaystyle R_{eq}=0.9\Omega\)

Notice that for resistors in parallel, the total resistance is never greater than the resistance of the smallest element.

Since the resistors \(\displaystyle R2\) and \(\displaystyle R3\) form a parallel network, removing \(\displaystyle R3\) from the circuit increases the resistance of that part of the circuit. Because the new circuit is the series combination of \(\displaystyle R1\) and \(\displaystyle R2\), the increased resistance leads to lower current in each of these resistors.