For any circuit, current $I$ flowing through the battery can be calculated as:

$$
begin{align}
I = dfrac{V} {R_e} tag{1}
end{align}
$$

where $V$ is voltage across the battery and $R_e$ is equivalent resistance of all the resistors in the circuit.
In our discussion we’ll be using a battery as a voltage source, but the principles apply to any voltage source.

Let’s take a look at a series circuit consisting of at least two resistors $R_1$ and $R_2$ and a voltage source $V$. Since resistors $R_1$ and $R_2$ are connected in series, same current $I$ will flow through each of the resistors and this same current will flow through the battery, stated as:

$$
I = I_1 = I_2
$$

We can calculate this current by applying the equation above as:

$$
I = dfrac{V}{R_e} = dfrac{V}{R_1 + R_2}
$$

where $R_e = R_1 + R_2$ is equivalent resistance of these resistors connected in series. Voltage across the resistors will add up to voltage across the battery, stated as:

$$
V = V_1 + V_2
$$

where $V_1$ is voltage across resistor $R_1$ and
$V_2$ is voltage across resistor $R_2$.

In short: In series connection of the resistors:

1) Same current $I$ flows through all of the resistors and the battery

$$
I = I_1 = I_2 = …
$$

2) Voltages across individual resistors add up to voltage $V$ across the battery

$$
V = V_1 + V_2 + …
$$

Let’s take a look at a parallel circuit consisting of at least two resistors $R_1$ and $R_2$ and a voltage source $V$. Since resistors $R_1$ and $R_2$ are connected in parallel, voltage $V$ across the battery will be equal to voltage $V_1$ across resistor $R_1$ and voltage $V_2$ across resistor $R_2$, stated as:

$$
V = V_1 = V_2
$$

Current $I$ flowing through the battery can be calculated by applying the first equation in this solution, as:

$$
begin{align*}
I &= dfrac{V}{R_e} \
I &= V dfrac{1}{R_e} tag{2}
end{align*}
$$

where $R_e$ is equivalent resistance of resistors $R_1$ and $R_2$ connected in parallel, which can be calculated from:

$$
dfrac{1}{R_e} = dfrac{1}{R_1} + dfrac{1}{R_2}
$$

We can plug in this expression for $dfrac{1}{R_e}$ into equation $(2)$ and have:

$$
begin{align*}
I &= V left( dfrac{1}{R_1} + dfrac{1}{R_2} right) \
tag{distribute the terms in the brackets} \
I &= dfrac{V}{R_1} + dfrac{V}{R_2} tag{3}
end{align*}
$$

Notice the Ohm’s law in the equation above. Ohm’s law states that current $I$ flowing through resistor $R$ with voltage $V$ across it is equal to:

$$
I = dfrac{V}{R}
$$

Given that voltage across all of the resistors in the parallel circuit is the same and equal $V$, we conclude that from equation $(3)$ it follows:

$$
begin{align*}
I &= dfrac{V}{R_1} + dfrac{V}{R_2} \
I &= I_1 + I_2
end{align*}
$$

This means that in parallel circuit, currents $I_1$ and $I_2$ flowing through resistors $R_1$ and $R_2$ add up to total current $I$ flowing through the battery.

In short: In parallel connection of the resistors:

1) Currents flowing through the individual resistors add up to current $I$ flowing through the battery:

$$
I = I_1 + I_2 + …
$$

2) Voltage across individual resistors is equal to voltage $V$ across the battery

$$
V = V_1 = V_2 = …
$$

In series connection of the resistors:
Voltages across individual resistors add up to voltage $V$ across the battery and same current $I$ flows through all of the resistors.

In parallel connection of the resistors:
Voltages across individual resistors are the same as voltage $V$ across the battery and currents through individual resistors add up to current $I$ flowing through the battery.

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We are given a parallel circuit with 4 resistors connected in parallel. Currents flowing through the 4 resistors are:

$$
begin{align*}
I_1 &= 120 ~mathrm{mA} = 0.120 ~mathrm{A} \
I_2 &= 250 ~mathrm{mA} = 0.250 ~mathrm{A} \
I_3 &= 380 ~mathrm{mA} = 0.380 ~mathrm{A} \
I_4 &= 2.1 ~mathrm{A} \
end{align*}
$$

For any circuit, current $I$ flowing through the source can be calculated as:

$$
begin{align}
I = dfrac{V} {R_e} tag{1}
end{align}
$$

where $V$ is voltage across the source and $R_e$ is equivalent resistance of all the resistors in the circuit.

Let’s take a look at a parallel circuit consisting of four resistors $R_1$, $R_2$, $R_3$ and $R_4$ and a voltage source $V$. Since resistors are connected in parallel, voltage $V$ across the source will be equal to voltage across each individual resistors:

$$
V = V_1 = V_2 = V_3 = V_4
$$

Current $I$ flowing through the source can be calculated by applying the first equation in this solution, as:

$$
begin{align*}
I &= dfrac{V}{R_e} \
I &= V dfrac{1}{R_e} tag{2}
end{align*}
$$

where $R_e$ is equivalent resistance of resistors these four resistors connected in parallel, which can be calculated from:

$$
dfrac{1}{R_e} = dfrac{1}{R_1} + dfrac{1}{R_2} + dfrac{1}{R_3} + dfrac{1}{R_4}
$$

We can plug in this expression for $dfrac{1}{R_e}$ into equation $(2)$ and have:

$$
begin{align*}
I &= V left( dfrac{1}{R_1} + dfrac{1}{R_2} + dfrac{1}{R_3} + dfrac{1}{R_4} right) \
tag{distribute the terms in the brackets} \
I &= dfrac{V}{R_1} + dfrac{V}{R_2} + dfrac{V}{R_3} + dfrac{V}{R_4} tag{3}
end{align*}
$$

Notice the Ohm’s law in the equation above. Ohm’s law states that current $I$ flowing through resistor $R$ with voltage $V$ across it is equal to:

$$
I = dfrac{V}{R}
$$

Given that voltage across all of the resistors in the parallel circuit is the same and equal $V$, we conclude that from equation $(3)$ it follows:

$$
begin{align*}
I &= dfrac{V}{R_1} + dfrac{V}{R_2} + dfrac{V}{R_3} + dfrac{V}{R_4} \
I &= I_1 + I_2 + I_3 + I_4 tag{4}
end{align*}
$$

This means that in parallel circuit, currents flowing through individual resistors add up to total current $I$ flowing through the source. We can now go back to equation $(4)$ to calculate the current $I$ supplied by the source:

$$
begin{align*}
I &= I_1 + I_2 + I_3 + I_4 \
I &= 0.120 ~mathrm{A} + 0.250 ~mathrm{A} + 0.380 ~mathrm{A} + 2.1 ~mathrm{A}
end{align*}
$$

$$
boxed{ I = 2.85 ~mathrm{A} }
$$

$$
I = 2.85 ~mathrm{A}
$$

Four resistors are connected in a series. Current through one of them is $I_1=810,,rm{mA}$. We need to determine the amount of current provided by the source.

When resistors are connected in a series, the current through them is **equal** and the very same. This means that the same current that goes through one of them, goes through **all of them**.
This also means that the current that goes through each of them also goes through the **source**. Finally this means that the current of the source is:
$$I_s=I_a$$
$$boxed{I_s=810,,rm{mA}}$$

$$I_s=810,,rm{mA}$$

In this problem we have a switch in a circuit consisting of a $P = 75 ~mathrm{W}$ bulb and a source with voltage $V= 120 ~mathrm{V}$.
$a)~~$
To find the potential difference (voltage) $V_s$ across the switch, let’s remember that the power rating $P$ of a device can be express in two different was:

$$
begin{align*}
P &= I^2 R \
P &= dfrac{V^2}{R}
end{align*}
$$

where $V$ is voltage across the device, $I$ is current flowing through the device and $R$ is resistance of the device. If we apply this to our switch, with voltage $V_s$ across it, current $I$ flowing through it and the rest of the series circuit and resistance $R_s$ we have:

$$
begin{align*}
P_s &= I^2 R_s tag{1}\
P_s &= dfrac{V_s^2}{R_s} tag{2}\
end{align*}
$$

To obtain the current $I$ in the circuit, we’ll apply Ohm’s law to the equivalent circuit, stated as:

$$
I = dfrac{V}{R_e}
$$

where $R_e$ is equivalent resistance of resistance of the bulb $R_b$ and resistance $R_s$ of the switch. Since $R_b$ and $R_s$ are connected in series, we have:

$$
R_e = R_b + R_s
$$

After we plug this into the equation above, we have:

$$
I = dfrac{V}{R_b + R_s}
$$

Note that the ideal switch is built so that it doesn’t affect the current $I$ in the circuit. From the equation above we conclude that this is only possible if resistance $R_s$ of the switch is zero:

$$
R_s = 0 ~~~rightarrow~~~ I = dfrac{V}{R_b}
$$

We aren’t interested in the current $I$ flowing through the circuit. In this case, since $R_s = 0$, we just want to see if this current is zero or non-zero. Since the switch is off, current will be flowing through the circuit and power output $P_s$ on the switch can be obtained from equation $(1)$ as:

$$
begin{align*}
P_s &= I^2 R_s \
tag{$R_s = 0$} \
P_s &= 0 \
tag{plug in $dfrac{V_s^2}{R_s} $ from equation $(2)$ } \
dfrac{V_s^2}{R_s} &= 0 \
V_s^2 &= 0 \
V_s &= 0
end{align*}
$$

As we can see, since resistance of the switch is taken as zero, voltage across the switch is zero:

$$
boxed{ V_s = 0 }
$$

$b)~~$ If we add another bulb, voltage across the switch will still stay the same and be equal zero. In any case in which resistance of the switch is taken as zero, $R_s = 0$ leads us to a zero-voltage switch. This can be obtained from Ohm’s law. Note that the Ohm’s law states that voltage $V$ across a device with resistance $R$ and current $I$ flowing through it is equal to:

$$
V= I R
$$

We see that if resistance of the device is taken as zero, voltage across it is also zero. This means that for an ideal switch, the following always applies:

$$
boxed{ V_s = 0 }
$$

Voltage $V_s$ across the switch is zero in both of the cases.

$$
V_s = 0
$$

$a)~~$ In this part of the problem we must find the current $I_{AB}$ flowing between points $A$ and $B$. We know that this current will be proportional to the potential difference between points $A$ and $B$. This potential difference $V_A – V_B$ can be calculated if we first find the voltage on each of the points $A$ and $B$.
Voltage $V_A$ at point $A$ is equal to the difference in voltage across the resistor to the left $V_{left, ~A}$ and resistor to the right $V_{right, ~A}$ from point $A$, stated as:

$$
V_A = V_{left, ~A} – V_{right, ~A}
$$

Voltages on these resistors can be obtained from Ohm’s law, which states that voltage $V$ across a resistor $R$ with current $I$ flowing through it is calculated as:
$V = I R$
Let $I_A$ be the current flowing through the branch with point $A$. This same current flows through both of the resistors $R$, from which we see:

$$
V_{left, ~A} = I_A R
$$

$$
V_{right, ~A} = I_A R
$$

We see that voltage at point $A$ is equal to:

$$
V_A = V_{left, ~A} – V_{right, ~A} = I_A R – I_A R = 0
$$

Analogously, we can find that voltage at point $B$ can be found as:

$$
V_B = V_{left, ~B} – V_{right, ~B} = I_B R – I_B R = 0
$$

where $V_{left, ~B}$ is voltage across the resistor on the left from point $B$ and $V_{right, ~B}$ is voltage across the resistor on the right from point $B$.
As said, current $I_{AB}$ is proportional to the potential difference between points $A$ and $B$, stated as:

$$
begin{align*}
I_{AB} propto (V_A – V_B) \
tag{$ V_A – V_B = 0 $} \
I_{AB} propto 0
end{align*}
$$

$$
boxed{ I_{AB} = 0 }
$$

In other words, there is no current in the wire connected between points $A$ and $B$.

$b)~~$ In this part of the problem we must see if current $I_A$ and $I_B$ flowing through the corresponding branches will change. Since potential difference between points $A$ and $B$ is zero, stated as

$$
V_A – V_B = 0
$$

we see that there is no current flowing between these two points and this circuit actually behaves in the same way as it did before we added another piece of wire between points $A$ and $B$ because there was no change in voltage between points $A$ and $B$.

$c)~~$ Current drawn from the battery won’t change either because circuit will behave in the same way as it did before we connected another piece of wire between points $A$ and $B$. This current $I$ can be calculated by dividing the voltage $V$ across the battery with equivalent resistance $R_e$ of all the resistors in the circuit.

$$
I = dfrac{V} {R_e}
$$

Equivalent resistance of the resistors in this circuit didn’t change since there was no change in current nor voltage across either of the resistors. As said, nothing changed in the circuit.

$d)~~$ Since there is no current between points $A$ and $B$, there will be no change in current through either of the resistors and thus voltage across either of the resistors won’t change. As said, since there is no potential difference between points $A$ and $B$, nothing will change in the circuit and there will be no change in voltage nor current anywhere in the circuit.

$a)~~ I_{AB} = 0$

$b)~~$ Current won’t change anywhere in the circuit

$c)~~$ Current drawn from the battery won’t change

$d)~~$ Voltage across either of the resistors won’t change.