We need to determine the rate of conversion of a lightbulb from electric energy to the light if the current is $I=0.5,,rm{A}$ and voltage is $V=125,,rm{V}$ with a $100,,%$ efficiency.

Power can be calculated as:
$$P=IV$$
Inserting values we get:
$$P=0.5cdot 125$$
Finally:
$$boxed{P=62.5,,rm{W}}$$

$$P=62.5,,rm{W}$$

In this problem we have a lamp with voltage $V= 12 ~mathrm{V}$ across it and current $I = 2 ~mathrm{A}$ flowing through it. To find the power $P$ used by the lamp, let’s remember that the power rating $P$ of a device with current $I$ flowing through it and voltage $V$ across it is calculated as:

$$
P = VI
$$

We’re given voltage $V$ across the lamp and current $I$ flowing through it, so we’ll plug in the given values into the equation above:

$$
begin{align*}
P &= VI \
tag{plug in the values} \
P &= 12 ~mathrm{V} cdot 2 ~mathrm{A}
end{align*}
$$

$$
boxed{ P = 24 ~mathrm{W} }
$$

$$
P = 24 ~mathrm{W}
$$

We need to determine the current flowing through a lightbulb of a power of $P=75,,rm{W}$ that is connected to an output with a voltage of $V=125,,rm{V}$.

In order to solve this problem, we are going to use an equation for electric power:
$$P=VI$$
From this equation we can extract current, $I$:
$$I=frac{P}{V}$$

Inserting values in the previous equation we get:
$$I=frac{75}{125}$$
Finally, the current is:
$$boxed{I=0.6,,rm{A}}$$

$$I=0.6,,rm{A}$$

We need to determine the amount of energy that was delivered to the starter in a period of $t=10,,rm{s}$ if the current throught the starter is equal to $I=210,,rm{A}$ and voltage is $V=12,,rm{v}$.

In order to solve this problem, first we need to calculate the **power output**.
We can do so using the next equation:
$$P=VI$$
Inserting values we get:
$$P=12cdot 210$$
$$P=2520,,rm{W}$$

Now we can calculate the energy delivered in a certain period. Power is equal to the energy exchanged over a period:
$$P=frac{E}{t}$$
This means that energy delivered is equal to the power acting for the mentioned period:
$$E=Pt$$
Inserting values we get:
$$E=2520cdot 10$$
Finally:
$$boxed{E=25200,,rm{J}}$$

$$E=25200,,rm{J}$$

In this problem we have a flashlight with voltage $V= 3 ~mathrm{V}$ across it. We’re told that the power rating $P$ of the flashlight bulb is $P = 0.9 ~mathrm{W}$. We must find the current $I$ flowing through this flashlight. To do this, let’s remember the power rating $P$ of a device with current $I$ flowing through it and voltage $V$ across it is calculated as:

$$
P = VI
$$

We can express the current $I$ flowing through this device as:

$$
I = dfrac{P}{V}
$$

We’re given power rating $P$ of the flashlight and voltage $V$ across the flashlight bulb, so we’ll plug in the values into the equation above:

$$
begin{align*}
I &= dfrac{P}{V} \
tag{plug in the values} \
I &= dfrac{ 0.9 ~mathrm{W} }{3 ~mathrm{V} }
end{align*}
$$

$$
boxed{ I = 0.3 ~mathrm{A} }
$$

$$
I = 0.3 ~mathrm{A}
$$

We need to determine the current through a panel lamp that has a resistance of $R=33,,rm{Omega}$ and is connected to the voltage of $V=12,,rm{V}$.

In order to solve this equation we need to use Ohm’s equation:
$$V=RI$$
From this we can get current:
$$I=frac{V}{R}$$

Inserting values:
$$I=frac{12}{33}$$
Finally:
$$boxed{I=0.3636,,rm{A}}$$

$$I=0.3636,,rm{A}$$

We need to determine the voltage of the source if the current in the circuit is $I=3.8,,rm{A}$ and the resistance is $R=32,,rm{Omega}$

In order to solve this problem we need to use Ohm’s equation:
$$I=frac{V}{R}$$
From this we can get voltage:
$$V=RI$$

Inserting values in the previous equation we get:
$$V=32cdot 3.8$$
Finally:
$$boxed{V=121.6,,rm{V}}$$

$$V=121.6,,rm{V}$$

We need to determine the resistance of the circuit if a voltage is $V=3,,rm{V}$ and current is $I=2cdot 10^{-4},,rm{A}$.

In order to solve this problem we need to use Ohm’s equation:
$$I=frac{V}{R}$$
From this we can get resistance:
$$R=frac{V}{I}$$

Inserting values in the previous equation:
$$R=frac{3}{2cdot 10^{-4}}$$
Finally, the resistance is:
$$boxed{R=15,,rm{kOmega}}$$

$$R=15,,rm{kOmega}$$

In this problem we have a lamp with current $I = 0.5 ~mathrm{A}$ flowing through it and voltage $V = 120 ~mathrm{V}$ across it. In first part of this problem we must find resistance $R$ of this lamp, while in second part we must find its power consumption $P$.

$a)~~$ Let’s remember that the Ohm’s law states that the current $I$ flowing through a device with resistance $R$ and voltage $V$ across it is equal to:

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

We can express resistance $R$ of this device from the equation above as:

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

Now that we have an equation that works in a general case, we can apply it for our lamp. Since we’re given voltage $V$ across the lamp and current $I$ flowing through it, we’ll simply plug in these values into the equation above:

$$
begin{align*}
R & = dfrac{V}{I} \
tag{plug in the values} \
R &= dfrac{120 ~mathrm{V} }{0.5 ~mathrm{A} }
end{align*}
$$

$$
boxed{ a)~~ R = 240 ~mathrm{Omega } }
$$

$b)~~$ To find the power consumption $P$ of this lamp, let’s remember that the power consumption $P$ of a device with voltage $V$ across it and current $I$ flowing through it is calculated as:

$$
P = V I
$$

Now that we have an equation that works in a general case, we can apply it for our lamp. Since we’re given voltage $V$ across the lamp and current $I$ flowing through it, we’ll simply plug in these values into the equation above:

$$
begin{align*}
P &= V I \
tag{plug in the values} \
P &= 120 ~mathrm{V} cdot 0.5 ~mathrm{A}
end{align*}
$$

$$
boxed{ b)~~ P = 60 ~mathrm{W } }
$$

$$
begin{align*}
a)~~ R &= 240 ~mathrm{Omega } \
b)~~ P &= 60 ~mathrm{W } \
end{align*}
$$

In this problem we have a lamp with power rating $P = 75 ~mathrm{W}$ and voltage $V = 125 ~mathrm{V}$ across it. In first part of this problem we must find current $I$ flowing through this lamp, while in second part we must find resistance $R$ of this lamp.

$a)~~$ To find the current $I$ flowing through this lamp, let’s remember that the power consumption $P$ of a device with voltage $V$ across it and current $I$ flowing through it is calculated as:

$$
P = V I
$$

We can express current $I$ flowing through the device as:

$$
I = dfrac{P}{V}
$$

Now that we have an equation that works in a general case, we can apply it for our lamp. Since we’re given voltage $V$ across the lamp
and power rating $P$ of the lamp, we’ll simply plug in these values into the equation above:

$$
begin{align*}
I &= dfrac{P}{V} \
tag{plug in the values} \
I &= dfrac{ 75 ~mathrm{W} }{125 ~mathrm{V} }
end{align*}
$$

$$
boxed{ a)~~ I = 0.6 ~mathrm{A} }
$$

$b)~~$ Let’s remember that the Ohm’s law states that the current $I$ flowing through a device with resistance $R$ and voltage $V$ across it is equal to:

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

We can express resistance $R$ of this device from the equation above as:

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

Now that we have an equation that works in a general case, we can apply it for our lamp. Since we’re given voltage $V$ across the lamp and current $I$ flowing through it, we’ll simply plug in these values into the equation above:

$$
begin{align*}
R & = dfrac{V}{I} \
tag{plug in the values} \
R &= dfrac{125 ~mathrm{V} }{0.6 ~mathrm{A} }
end{align*}
$$

$$
boxed{ b)~~ R = 208.3333 ~mathrm{Omega } }
$$

$$
begin{align*}
a)~~ I &= 0.6 ~mathrm{A} \
b)~~ R &= 208.3333 ~mathrm{Omega } \
end{align*}
$$

In this problem we have a circuit consisting of a battery with voltage $V = 125 ~mathrm{V}$ and a lamp with power rating $P_{i} = 75 ~mathrm{W}$ and resistance $R_{lamp} = 208.3333 ~mathrm{Omega}$. Current supplied by the battery is $I_i = 0.6 ~mathrm{A}$. These are the initial conditions of the circuit (thus the index “i”). We’ll now add an additional resistor $R_{added}$ to reduce the current to half the original value, which means that current in this final case is

$$
I_f = dfrac{I_i}{2} = 0.3 ~mathrm{A}
$$

We need to find the voltage $V_{f}$ across the lamp,

resistance $R_{added}$ added to the circuit and

power $p_{f}$ dissipated on the lamp.
Note that we used index “f” to denote the final case.

$a)~~$ To find the potential difference $V_f$ across the lamp, let’s remember that the Ohm’s law states that the current $I$ flowing through a device with resistance $R$ and voltage $V$ across it is given as:

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

We can express the voltage $V$ across the device as:

$$
V = I R
$$

Now that we have an equation for finding voltage $V$ across a device that works in general case, we can apply it for our lamp in the final case. Voltage across the lamp is $V_f$, while current flowing through it is $I_f$ and resistance of the lamp, of course, didn’t change. After we apply the equation above to this lamp, we have:

$$
begin{align*}
V_f &= I_f R \
tag{plug in the values} \
V_f &= 0.3 ~mathrm{A} cdot 208.3333 ~mathrm{Omega}
end{align*}
$$

$$
boxed{a)~~ V_f = 62.5 ~mathrm{V} }
$$

$b)~~$ To find the additional resistance $R_{added}$ added to the circuit, let’s remember that the voltage drops across individual resistors must add up to voltage $V$ across the battery. This means:

$$
V_f + V_{added} = V
$$

where $V_f$ is voltage voltage across the lamp, calculated in part $a)$ and $V_{added}$ is voltage across the additional resistor.
We can express voltage $V_{added}$ across the additional resistor as:

$$
V_{added} = V – V_f
$$

Remember that from the Ohm’s law it follows that resistance $R$ of the resistor can be calculated as:

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

where $V$ is voltage across this particular resistor and $I$ is current flowing through it. This means that resistance $R_{added}$ of the additional resistor, with voltage $V_{added}$ across it and current $I_f$ flowing through it can be calculated as:

$$
begin{align*}
R_{added} &= dfrac{V_{added}}{I_f} \
tag{plug in $V_{added} = V – V_f $} \
R_{added} &= dfrac{ V – V_f }{I_f} \
tag{plug in the values} \
R_{added} &= dfrac{ 125 ~mathrm{V} – 62.5 ~mathrm{V} }{0.3 ~mathrm{A} }
end{align*}
$$

$$
boxed{ R_{added} = 208.3333 ~mathrm{Omega} }
$$

As expected, current dropped to half of its initial value because another resistor, with resistance equal to resistance of the lamp was connected in series to the lamp.

$c)~~$ To find the power $P_f$ dissipated on this lamp, let’s remember that the power consumption $P$ of a device with voltage $V$ across it and current $I$ flowing through it is calculated as:

$$
P = V I
$$

Now that we have an equation that works in a general case, we can apply it for our lamp. Since we’ve calculated voltage $V_f$ across the lamp in final case and current $I_f$ flowing through it, we’ll simply plug in these values into the equation above:

$$
begin{align*}
P_f &= V_f I_f \
tag{plug in the values} \
P &= 62.5 ~mathrm{V} cdot 0.3 ~mathrm{A}
end{align*}
$$

$$
boxed{ c)~~ P = 18.75 ~mathrm{W } }
$$

$$
begin{align*}
& a)~~ V_f = 62.5 ~mathrm{V} \
& b)~~ R_{added} = 208.3333 ~mathrm{Omega} \
& c)~~ P = 18.75 ~mathrm{W }
end{align*}
$$

We are told to draw a circuit diagram that includes a
$V= 60 ~mathrm{V}$ battery
, an ammeter
and a resistance $R = 12.5 ~mathrm{Omega }$
all connected in series.
We must also indicate the reading on ammeter and direction of the current $I$.

Reading on ammeter connected in series to the resistor $R$ is the same as current $I$ flowing through the resistor $R$.
From Ohm’s law we know that current $I$ flowing through resistor $R$ with voltage $V$ across it is equal to:

$$
begin{align*}
I &= dfrac{V}{R} \
I &= dfrac{ 60 ~mathrm{V} }{12.5 ~mathrm{Omega } }\
I &= 4.8 ~mathrm{A}
end{align*}
$$

Now that we know the current $I$ flowing through the circuit, and knowing that direction of the current is from positive to negative terminal of the battery, we can now draw the circuit diagram.

Hint: Apply Ohm’s law to the circuit and express $R$ from the equation. Click for further explanation.

$$
I = 4.8 ~mathrm{A}
$$

We are told to draw a circuit diagram that includes a
$V= 4.5 ~mathrm{V}$ battery
, an ammeter that reads $I = 85 ~mathrm{mA}$
and a resistor $R$
all connected in series.
We must also indicate the positive terminal on the battery and choose a conventional direction of current $I$ through the circuit.

Reading on ammeter connected in series to the resistor $R$ is the same as current $I$ flowing through the resistor $R$.
From Ohm’s law we know that current $I$ flowing through resistor $R$ with voltage $V$ across it is equal to:

$$
begin{align*}
I &= dfrac{V}{R} \
tag{express $R$ from the equation above} \
R &= dfrac{V}{I} \
R &= dfrac{ 4.5 ~mathrm{V} }{ 85 ~mathrm{mA} }\
R &= 52.9 ~mathrm{Omega }
end{align*}
$$

We are given current $I$ flowing through the circuit. We that positive terminal of the battery is the one labelled with $+$ sign. We also know that battery can be indicated as two parallel lines, normal to the wires in the circuit, where one of the lines is longer than the other. The longer line is positive terminal of the battery, while the shorter line is negative terminal of the battery. Direction of the current is from positive to negative terminal of the battery. We can now draw the circuit diagram.