Here we analyze what happens if we exchange one of the four resistors that are connected in series inside the circuit with the ideal wire. Since for resistors in series, it is known:

$$
R_{mathrm{eq}}=R_{1}+R_{2}+R_{3}+R_{4} = 4R
$$

If we disconnect one, the total (equal) resistance will be only less then for one resistor and equal to the sum of those that are left:

$$
R_{mathrm{eq}}=R_{1}+R_{2}+R_{3}=3R
$$

$$
R_{mathrm{eq}}=3R
$$

Suppose $R$ is the resistance of the each resistor.

So when two of them connected parallel, the equivalent resistance will be

$$
R’=left[frac{1}{R}+frac{1}{R}right]^{-1}=frac{R}{2}
$$

When the third resistance will be connected in parallel with the above resistance then the equivalent resistance will be

$$
R”=left[frac{1}{R}+frac{1}{R}+frac{1}{R}right]^{-1}=frac{R}{3}
$$

Hence we have

$$
frac{R”}{R’}=frac{R/3}{R/2}=frac{2}{3}
$$

So final equivalent resistance will be $2/3$ of the initial equivalent resistance.

The car headlight is connected in $textbf{parallel}$. We know this from everyday experience, as we can often see a car driving only with one light (left or right). If they would be connected in a series non of it would work, because if the current can not go through one light, it can not go through any other light.

Since they are connected in parallel even if one is not working, the other can work normally. We can also relate this we decorative light (e.g. for Christmas), they are all connected in parallel, so if few are not working, others can work normally. If they would be connected in series, breaking only one would mean stopping the work of all others, as well.

Here we analyze the problem in the picture below, we know that the current through the first resistor is 0.72 A. Since the second one is connected in series, we know that the current is also the same through it.

We need to calculate the emf of battery and we can do that using Ohm’s law:

$$
I=frac{varepsilon}{R_{mathrm{eq}}} Rightarrow varepsilon= IR_{eq}
$$

We need to first calculate equal resistance since they are connected in series we have:

$$
R_{mathrm{eq}}=R_{1}+R_{2}
$$

Putting in the numbers we have:

$$
R_{mathrm{eq}}=89 Omega + 130 Omega
$$

which gives the result of:

$$
R_{mathrm{eq}}=219 Omega
$$

Now the emf of the battery is going to be:

$$
varepsilon=I R_{e q} = 0.72 mathrm{A} cdot 219 Omega
$$

which gives the result of:

$$
varepsilon=I R_{e q} = boxed{color{#c34632}157.68 mathrm{V}}
$$

$$
varepsilon = 157.68 mathrm{V}
$$

Since the resistors are in parallel we do know they are on the same potential difference, so both resistors are having a potential difference of 7.7 V. When it comes to the current, we can see that some amount of current is given from the battery and then it $textbf{separates}$ on the node with the resistor one. We calculate the equal resistance and by Ohm’s law, we calculate the current.

We know that when resistors are in parallel equal resistance is:

$$
frac{1}{R_{mathrm{eq}}}=frac{1}{R_{1}}+frac{1}{R_{2}}
$$

Putting in the numbers we have:

$$
frac{1}{R_{mathrm{eq}}}=frac{1}{210 Omega}+frac{1}{130 Omega} = frac{1}{0.01245}
$$

We have te result of:

$$
R_{mathrm{eq}} = 80.3 Omega
$$

From Ohm’s law we know:

$$
I=frac{V}{R_{text {eq }}} = frac{7.7 mathrm{V}}{80.3 Omega}
$$

We have the result of:

$$
boxed{color{#c34632}I=0.096 mathrm{A}}
$$