We need to determine the equivalent resistance of a circuit shown in a diagram.
There are three resistors with resistances $R_1=3,,rm{Omega}$, $R_2=12,,rm{Omega}$ and $R_3=4,,rm{Omega}$ connected in parallel.

In order to solve this problem, we need to use equation for equivalent resistance of resistors in a parallel conection:
$$frac{1}{R_e}=frac{1}{R_1}+frac{1}{R_2}+frac{1}{R_3}$$

Inserting given values into the previous equation we will get:
$$frac{1}{R_e}=frac{1}{3}+frac{1}{12}+frac{1}{4}$$
Solving further we get:
$$frac{1}{R_e}=frac{8}{12}$$
Which finally means that:
$$R_e=frac{12}{8}$$

Previous result can also be expressed as:
$$boxed{R_e=1.5,,rm{Omega}}$$
Which is **answer C**.

In this problem we have three resistors with resistances
$R_1 = 3 ~mathrm{Omega }$,
$R_2 = 12 ~mathrm{Omega }$ and
$R_3 = 4 ~mathrm{Omega }$ connected in parallel.
To find the current $I$ in the circuit, we’ll use the following equation:

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

where $V = 6 ~mathrm{V}$ is voltage across the battery and $R_e$ is equivalent resistance of all resistors in the circuit. We see that in order to calculate the current $I$ supplied from the battery, we first need to calculate the equivalent resistance $R_e$ of resistors in the circuit.

As we can see, the three resistors are connected in parallel.
To calculate the equivalent resistance of resistors connected in parallel, we use a following equation:

$$
begin{align*}
dfrac{1}{R_e} &= dfrac{1}{R_1} + dfrac{1}{R_2} + dfrac{1}{R_3} \
tag{plug in the values} \
dfrac{1}{R_e} &= dfrac{ 1 }{3 ~mathrm{Omega } } + dfrac{1}{12 ~mathrm{Omega } } + dfrac{1}{4 ~mathrm{Omega } } \
tag{common denominator is $12 ~mathrm {Omega } $} \
dfrac{1}{R_e} &= dfrac{4}{12 ~mathrm {Omega } } + dfrac{1}{12 ~mathrm {Omega } } + dfrac{3}{12 ~mathrm {Omega } } \
dfrac{1}{R_e} &= dfrac{8}{12 ~mathrm{Omega }} \
tag{find reciprocal value of the equation above} \
R_e &= dfrac{12}{8} ~mathrm{Omega } \
R_e &= 1.5 ~mathrm{Omega }
end{align*}
$$

To calculate the current $I$ flowing in the circuit, we’ll plug in the calculated value for equivalent resistance $R_e$ to equation (1):

$$
begin{align*}
I &= dfrac{V}{R_e} \
tag{plug in the values} \
I &= dfrac{6 ~mathrm{V}}{1.5 ~mathrm{Omega }}
end{align*}
$$

$$
D)~~ I = 4 ~mathrm{A }
$$

$$
D)~~ I = 4 ~mathrm{A }
$$

We need to use the given scheme in order to calculate the current in the resistor $R_3=4,,rm{Omega}$ is the resistances in remaining two resistors are: $R_1=3,,rm{Omega}$ and $R_2=12,,rm{Omega}$ and the supply voltage is $epsilon =6,,rm{V}$. Resistors are connected in parallel.

First we need to notice that the resistors are connected in parallel. This means that the voltage drop on them is equal inbetween them and equal to the voltage of the battery:
$$V_3=epsilon =6,,rm{V}$$

Now we can use Ohm’s Law in order to calculate current since we know both the voltage drop and resistance of the resistor number $3$:
$$I_3=frac{V_3}{R_3}$$

Inserting values into the previous equation we get:
$$I_3=frac{6}{4}$$
Finally, the current is:
$$boxed{I_3=1.5,,rm{A}}$$
Which is **answer B**.

We need to use the given scheme in order to calculate the voltage drop in the resistor $R_2=12,,rm{Omega}$ is the resistances in remaining two resistors are: $R_1=3,,rm{Omega}$ and $R_3=4,,rm{Omega}$ and the supply voltage is $epsilon =6,,rm{V}$. Resistors are connected in parallel.

First thing we need to notice is that the resistors are connected into parallel.
This means that the voltage on them is equal inbetween them.
$$V=V_1=V_2=V_3$$

Since the parallel connection is directly connected to the supply voltage, the voltage drop of the connection needs to be **equal to the supply voltage**
$$V=epsilon$$

Combining two previous equations we get:
$$V_2=epsilon$$

Inserting given values into the equation we get the voltage drop on the second resistor:
$$boxed{V_2=6,,rm{V}}$$
Which is **answer D**.

In this problem we have a series-parallel circuit consisting of resistors $R_B = 25 ~mathrm{Omega }$ and $R_C = 15 ~mathrm{Omega }$ connected in parallel, while resistor $R_A = 12 ~mathrm{Omega }$ is connected in series series to parallel combination of resistors $R_B$ and $R_C$. To find the equivalent resistance, notice that resistors $R_B$ and $R_C$ can be changed for resistor $R_{BC}$ and that this resistor $R_{BC}$ is actually equivalent resistor of parallel connection of resistors $R_B$ and $R_C$. If we change resistors $R_B$ and $R_C$ for their equivalent resistor $R_{BC}$ we’d have resistors $R_A$ and $R_{BC}$ connected in series. To find the equivalent resistance $R_e$ of all resistors in the circuit, we’ll finally find equivalent resistance of resistors $R_A$ and $R_{BC}$ connected in series.

As said, resistors $R_B$ and $R_C$ are connected in parallel. To find their equivalent resistance $R_{BC}$ , we’ll use a following equation:

$$
begin{align*}
dfrac{1}{R_{BC}} &= dfrac{1}{R_B} + dfrac{1}{R_C} \
tag{common denominator is $R_B R_C$} \
dfrac{1}{R_{BC}} &= dfrac{R_C + R_B}{R_B R_C } \
tag{find reciprocal value of the equation above} \
R_{BC} &= dfrac{R_B R_C }{R_C + R_B} \
tag{plug in the values} \
R_{BC} &= dfrac{ 25 ~mathrm{Omega } cdot 15 ~mathrm{Omega } }{25 ~mathrm{Omega } + 15 ~mathrm{Omega } } \
R_{BC} &= 9.735 ~mathrm{Omega }
end{align*}
$$

As said, resistor $R_{BC}$ is connected in series to resistor $R_A$, which means that equivalent resistance $R_e$ of all resistors in the circuit is actually equivalent resistance of resistors $R_{BC}$ and $R_A$ connected in series. We know that equivalent resistance $R_e$ of resistors connected in series is equal to a sum of resistances of individual resistors:

$$
begin{align*}
R_e &= R_{BC} + R_A \
tag{plug in the values} \
R_e &= 9.735 ~mathrm{Omega } + 12 ~mathrm{Omega } \
R_e &= 21.375 ~mathrm{Omega }
end{align*}
$$

We found that equivalent resistance $R_e$ is approximately equal to $21.4 ~mathrm{Omega }$. Final answer is:

$$
boxed{ C)~~ R_e approx 21.4 ~mathrm{Omega } }
$$

$$
C)~~ R_e approx 21.4 ~mathrm{Omega }
$$

In this problem we have a series-parallel circuit consisting of resistors $R_B = 25 ~mathrm{Omega }$ and $R_C = 15 ~mathrm{Omega }$ connected in parallel, while resistor $R_A = 12 ~mathrm{Omega }$ is connected in series series to parallel combination of resistors $R_B$ and $R_C$.
To find the current $I$ in the circuit, we’ll use the following equation:

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

where $V = 60 ~mathrm{V}$ is voltage across the battery and $R_e$ is equivalent resistance of all resistors in the circuit. We see that in order to calculate the current $I$ supplied from the battery, we first need to calculate the equivalent resistance $R_e$ of resistors in the circuit.
To find the equivalent resistance, notice that resistors $R_B$ and $R_C$ can be changed for resistor $R_{BC}$ and that this resistor $R_{BC}$ is actually equivalent resistor of parallel connection of resistors $R_B$ and $R_C$. If we change resistors $R_B$ and $R_C$ for their equivalent resistor $R_{BC}$ we’d have resistors $R_A$ and $R_{BC}$ connected in series. To find the equivalent resistance $R_e$ of all resistors in the circuit, we’ll finally find equivalent resistance of resistors $R_A$ and $R_{BC}$ connected in series.

As said, resistors $R_B$ and $R_C$ are connected in parallel. To find their equivalent resistance $R_{BC}$ , we’ll use a following equation:

$$
begin{align*}
dfrac{1}{R_{BC}} &= dfrac{1}{R_B} + dfrac{1}{R_C} \
tag{common denominator is $R_B R_C$} \
dfrac{1}{R_{BC}} &= dfrac{R_C + R_B}{R_B R_C } \
tag{find reciprocal value of the equation above} \
R_{BC} &= dfrac{R_B R_C }{R_C + R_B} \
tag{plug in the values} \
R_{BC} &= dfrac{ 25 ~mathrm{Omega } cdot 15 ~mathrm{Omega } }{25 ~mathrm{Omega } + 15 ~mathrm{Omega } } \
R_{BC} &= 9.735 ~mathrm{Omega }
end{align*}
$$

As said, resistor $R_{BC}$ is connected in series to resistor $R_A$, which means that equivalent resistance $R_e$ of all resistors in the circuit is actually equivalent resistance of resistors $R_{BC}$ and $R_A$ connected in series. We know that equivalent resistance $R_e$ of resistors connected in series is equal to a sum of resistances of individual resistors:

$$
begin{align*}
R_e &= R_{BC} + R_A \
tag{plug in the values} \
R_e &= 9.735 ~mathrm{Omega } + 12 ~mathrm{Omega } \
R_e &= 21.375 ~mathrm{Omega }
end{align*}
$$

To find the current $I$ flowing in the circuit, we’ll apply equation (1) to our circuit. Equation (1) states:

$$
begin{align*}
I &= dfrac{V}{R_e} \
I &= dfrac{60 ~mathrm{V}}{ 21.375 ~mathrm{Omega } } \
I &= 2.807 ~mathrm{A}
end{align*}
$$

We found that current $I$ supplied to the circuit is approximately equal to $2.8 ~mathrm{A}$. Final answer is:

$$
boxed{ C)~~ I approx 2.8 ~mathrm{A } }
$$

$$
C)~~ I approx 2.8 ~mathrm{A }
$$

We need to calculate the total resistance of the circuit in which there are $8$ resistors with a resistance of $R=12,,rm{Omega}$ connected in series.

Equivalent resistance of a series connection can be calculated as a sum of all of the resistances in that series connection:
$$R_e=R+R+R+…$$

Since we have $8$ of them, we can write:
$$R_e=8cdot R$$

Inserting given values into the equation:
$$R_e=8cdot 12$$
Finally, the total resistance is:
$$boxed{R_e=96,,rm{Omega}}$$
Which means that correct **answer is D**.

In this problem we are given 4 statements and we must conclude which one of the statements is true. Statements are:

$A)~~$ The resistance of a typical ammeter is
very high.

$B)~~$ The resistance of a typical voltmeter is
very low.

$C)~~$ Ammeters have zero resistance.

$D)~~$ A voltmeter causes a small change
in current.

Let’s examine statement $A)$. Notice that ammeter is connected in series to the resistor with arbitrary resistance $R$. If there was current $I$ flowing through this resistor and we connected an ammeter with presumably high resistance, resistance in that part of the circuit will be equal to $R_A + R$, where $R_A$ is resistance of the ammeter. Since resistance is now much higher than it was, current will decrease by a lot and measured current will be much different compared to current $I$ that we were supposed to measure. In fact, resistance of an ideal ammeter is actually equal to zero (but never truly zero). We conclude that statement $A)$ is false.

Let’s examine statement $B)$. If we want to measure voltage $V$ across a certain resistor $R$, we’ll connect a voltmeter in parallel to this resistor. Let’s assume that we did connected a voltmeter with very low resistance $R_V$ in parallel to resistor $R$ and let’s find equivalent resistance $R_e$ of parallel connection of resistor $R$ and resistor $R_V$ in voltmeter:

$$
begin{align*}
dfrac{1}{R_e} &= dfrac{1}{R} + dfrac{1}{R_V} \
tag{common denominator is $R R_V$} \
dfrac{1}{R_e} &= dfrac{R_V + R}{R_V R} \
tag{find reciprocal value of the equation above} \
R_e &= dfrac{R_V R}{R_V + R} \
tag{divide numerator and denominator by $R$} \
R_e &= dfrac{R_V}{ dfrac{R_V + R}{R}} \
R_e &= dfrac{R_V}{ 1 + dfrac{R_V}{R} } \
tag{since we assume $R_V$ is very low, $dfrac{R_V}{R} << 1$} \
R_e & approx dfrac{R_V}{1 + 0 } \
R_e & approx R_V
end{align*}
$$

As we can see, if voltmeter had a very low resistance, connecting the voltmeter in parallel would result in very low equivalent resistance of all resistors in the circuit, which in turn increases the current flowing in the circuit. This current will be high enough to burn out this voltmeter. Since we can usually measure the voltage across a device without voltmeter dying, we conclude that statement $B)$ is also false. This is due to a fact that voltage of an ideal voltmeter is actually infinite (but never truly infinite).

$C)~~$ As stated above, resistance of an ideal ammeter is zero, but never truly zero. The reason for this is due to a fact that any wire inside of ammeter has some low, negligible but non-zero resistance. Keep in mind that resistance $R_A$ of ammeter is low and what we described above in part $B)$ is precisely what happens when we connect the ammeter in parallel to a resistor – high current flows through the ammeter and ammeter burns out. Since resistance of the ammeter is not truly zero, we conclude that statement $C)$ is also false.

$$
D)~~
$$

If we connect the voltmeter in parallel to resistor $R$, resistance in that part of the circuit will change from resistance of resistor $R$ to equivalent resistance $R_e$ of parallel combination of resistor $R$ and
resistor $R_V$ in the voltmeter. Keep in mind that resistance $R_V$ of voltmeter is actually really high, which means:

$$
R_V >> R
$$

As described above, equivalent resistance $R_e$ of resistor $R$ and resistor $R_V$ connected in parallel is equal to:

$$
begin{align*}
R_e &= dfrac{R_V R}{R_V + R} \
tag{divide the numerator and denominator by $R_V$} \
R_e &= dfrac{ R }{dfrac{R_V + R}{R} } \
R_e &= dfrac{ R }{1 + dfrac{ R}{R_V} } \
tag{ let $dfrac{ R}{R_V} = alpha $ } \
R_e &= dfrac{R}{1 + alpha }
end{align*}
$$

Notice that, as said, resistance $R_V$ is very high, but not infinite. Thus term $dfrac{R}{R_V} = alpha$ is very small, but non-zero.
Current $I$ flowing through part of the circuit where we connected the voltmeter will be equal to:

$$
begin{align*}
I &= dfrac{V}{R_e} \
tag{plug in $R_e = dfrac{R}{1 + alpha }$}
I &= dfrac{V}{ dfrac{R}{1 + alpha } } \
I &= dfrac{V}{R} (1 + alpha)
end{align*}
$$

where $V$ is voltage across this part of the circuit.
Current $I_0$ flowing through the circuit when voltmeter is not connected is equal to:

$$
I_0 = dfrac{V}{R}
$$

Now let’s find the difference between the current $I_0$ flowing through this part of the circuit when voltmeter is not connected to it and current $I$ flowing through this part of the circuit when we connect the voltmeter:

$$
begin{align*}
I – I_0 &= dfrac{V}{R} (1 + alpha) – dfrac{V}{R} \
I – I_0 &= dfrac{V}{R} + dfrac{V}{R} alpha – dfrac{V}{R} \
I – I_0 &= dfrac{V}{R} alpha \
tag{plug in $alpha = dfrac{R }{R_V} $ } \
I – I_0 &= dfrac{V}{R} dfrac{R}{R_V} \
tag{cancel out $R$} \
I – I_0 &= dfrac{V}{R_V}
end{align*}
$$

Notice that this change in current is equal to $dfrac{V}{R_V}$. Since voltmeter is connected in parallel to this part of the circuit, voltage $V$ is also voltage across the voltmeter and since $R_V$ is its resistance, we conclude that this difference actually represents the current $I_V$ flowing through the voltmeter. Now that we have proven that voltmeter causes a small change in current in the circuit, we conclude that statement $D)$ is true.

$$
boxed{ text{Statement $D)$ is true } }
$$

$$
text{Statement $D)$ is true }
$$

In this problem we have 15 identical outdoor lamps connected to a $V = 12 ~mathrm{V}$ battery of a car. When we connected 15 lamps, they didn’t glow. Ammeter showed that current through the lamps is
$I_i = 0.35 ~mathrm{A}$, while current of at least $I_f = 0.5 ~mathrm{A}$ is required for lamps to work. Since lamps are connected in series and equivalent resistance increases the more resistors we add to the circuit, we’ve concluded that we connected to many lamps and because of this total resistance of the lamps is too high and current flowing through them doesn’t reach the value of $0.5 ~mathrm{A}$ needed for lamps to glow. Current $I$ flowing through a circuit can be calculated as:

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

where $V$ is voltage across the voltage source and $R_e$ is equivalent resistance of all resistors in the circuit. We see that current of $I_f = 0.5 ~mathrm{V}$ can be calculated as:

$$
I_f = dfrac{V}{R_{ef}}
$$

where $R_{ef}$ is equivalent resistance of the lamps in the final case, after we remove some of the lamps. We can express equivalent resistance $R_e$ in a circuit from the equation above as:

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

If we connect 15 lamps to the car battery with voltage $V$, we have the initial value of the equivalent resistance $R_{ei}$. In this case current $I_i$ flows in the circuit. We must also keep in mind that in first case we have 15 lamps connected in series, which means that equivalent resistance $R_{ei}$ in the initial case is equal to:

$$
R_{ei} = 15 R
$$

where $R$ is resistance of each of the lamps.

Let’s apply equation (2) to initial equivalent resistance $R_{ei}$:

$$
begin{align*}
R_{ei} &= dfrac{V}{I_i} \
tag{plug in $R_{ei} = 15 R$} \
15 R &= dfrac{V}{I_i} \
tag{express $R$ from the equation above} \
R &= dfrac{V}{15 I_i} \
tag{plug in the values} \
R &= dfrac{ 12 ~mathrm{V} }{15 cdot 0.35 ~mathrm{A} }
end{align*}
$$

$$
boxed{ R = 2.2857 ~mathrm{Omega } }
$$

Now that we’ve found the resistance across of each of the lamps, let’s consider the following. If we remove some lamps, we’ll have $N$ lamps left in the circuit. In this case, since lamps are connected in series, equivalent resistance $R_{ef}$ in final case will be equal to:

$$
R_{ef} = N R
$$

Now let’s apply equation (1) to current $I_f$ in this final case of the circuit, where we removed some of the lamps:

$$
begin{align*}
I_f &= dfrac{V}{R_{ef}} \
tag{plug in $R_{ef} = NR $} \
I_f &= dfrac{V}{NR } \
tag{multiply by $NR$} \
I_f NR &= V \
tag{express $N$ from the equation above} \
N &= dfrac{V}{I_f R} \
tag{plug in the values} \
N &= dfrac{ 12 ~mathrm{V}}{ 0.5 ~mathrm{A} cdot 2.2857 ~mathrm{Omega } } \
N &= 10.5
end{align*}
$$

$$
boxed{ N < 10 }
$$

We found that the maximum number of lamps that we can have in a circuit in order for current through the circuit to be at least $0.5 ~mathrm{A}$ is 10. Notice that we can't have a non-integer number of lamps and that if we had $10.5$ lamps in the circuit, current in the circuit would be exactly equal to $0.5 ~mathrm{A}$. If we had any higher number of lamps in the circuit, current would be lower than this value and still not enough for lamps to light up.
Now that we know that the maximum number of lamps in the circuit is 10, while initial number of lamps in the circuit was 15, we conclude that

$$
boxed{ text{ We had to remove 5 lamps for lamps to light up. } }
$$

$$
text{ We had to remove 5 lamps for lamps to light up. }
$$

In this problem we have a battery with voltage $V = 8 ~mathrm{V}$ and four resistors with resistances $R_1 = 4 ~mathrm{Omega }$, $R_2 = 8 ~mathrm{Omega }$, $R_3 = 13 ~mathrm{Omega }$ and $R_4 = 15 ~mathrm{Omega }$ connected to it in series. We must find the current $I$ flowing in the circuit and power output $P$ of the circuit.
Current $I$ flowing through a circuit can be calculated as:

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

where $V$ is voltage across the voltage source and $R_e$ is equivalent resistance of all resistors in the circuit, while total power output $P$ in the circuit can be found as:

$$
begin{align*}
P &= V I tag{2}
end{align*}
$$

where $V$ is voltage across the source and $I$ is current supplied from the source.

Since these resistors are connected in series, equivalent resistance of these resistors is found by adding up the resistance of individual resistors:

$$
begin{align*}
R_e &= R_1 + R_2 + R_3 + R_4 \
tag{plug in the values} \
R_e &= 4 ~mathrm{Omega } + 8 ~mathrm{Omega } + 13 ~mathrm{Omega } + 15 ~mathrm{Omega }
end{align*}
$$

$$
boxed{ R_e = 40 ~mathrm{Omega } }
$$

We can find current $I$ flowing in the circuit by plugging in the values in equation (1):

$$
begin{align*}
I &= dfrac{V}{R_e} \
tag{plug in the values} \
I &= dfrac{8 ~mathrm{V}}{ 40 ~mathrm{Omega } }
end{align*}
$$

$$
boxed{ I = 0.2 ~mathrm{A} }
$$

Note that since resistors are connected in series, same current $I$ calculated above flows through each of the resistors, and it is the same current $I$ supplied from the battery.

We can find total power output $P$ on the resistors by plugging in the values in equation $(2)$:

$$
begin{align*}
P &= I V \
tag{plug in the values} \
P &= 0.2 ~mathrm{A} cdot 8 ~mathrm{V}
end{align*}
$$

$$
boxed{ P= 1.6 ~mathrm{W} }
$$

$$
I = 0.2 ~mathrm{A}
$$

$$
P= 1.6 ~mathrm{W}
$$