The resistance of a 120 V light bulb that draws a current of 1.25 A is .

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4. The resistance of a 120 V light bulb that draws a current of 1.25 A is ________.

According to Ohm's Law, the resistance $R$ is given by
$R = \frac{V}{I}$
where $V$ is the voltage and $I$ is the current.
Substitute the given values
$V = 120 \, \text{V}$ and $I = 1.25 \, \text{A}$ .
$R = \frac{120 \, \text{V}}{1.25 \, \text{A}}$ $R = 96 \, \Omega$
The correct option is B.
$\boxed{96 \, \Omega}$
5. A certain wire has a resistance R. Another wire, exactly the same except having twice the diameter, has the resistance ________.
The resistance $R$ of a wire is given by the formula
$R = \rho \frac{L}{A}$
where $\rho$ is the resistivity of the material, $L$ is the length, and $A$ is the cross-sectional area.
The cross-sectional area $A$ of a circular wire is given by $A = \pi r^2$ , where $r$ is the radius.
Since diameter $d = 2r$ , then $r = d/2$ , so $A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$ .
Substitute the area into the resistance formula
$R = \rho \frac{L}{\frac{\pi d^2}{4}} = \frac{4 \rho L}{\pi d^2}$
For the second wire, the diameter is $d' = 2d$ .
All other parameters ( $\rho$ , $L$ ) are the same. Let its resistance be $R'$ .
$R' = \frac{4 \rho L}{\pi (d')^2} = \frac{4 \rho L}{\pi (2d)^2} = \frac{4 \rho L}{\pi (4d^2)} = \frac{1}{4} \left( \frac{4 \rho L}{\pi d^2} \right)$
From Step 3, we know that $R = \frac{4 \rho L}{\pi d^2}$ .
So,
$R' = \frac{1}{4} R$
The correct option is D.
$\boxed{\frac{1}{4}R}$
6. The current in the 5 Ω resistor is ________.
The resistors $5 \, \Omega$ and $10 \, \Omega$ are connected in series.
The total equivalent resistance $R_{eq}$ for series resistors is the sum of individual resistances:
$R_{eq} = R_1 + R_2$ $R_{eq} = 5 \, \Omega + 10 \, \Omega = 15 \, \Omega$
A potential difference $V = 60 \, \text{V}$ is applied across the series combination.
According to Ohm's Law, the total current $I$ flowing through the circuit is:
$I = \frac{V}{R_{eq}}$ $I = \frac{60 \, \text{V}}{15 \, \Omega}$ $I = 4 \, \text{A}$
In a series circuit, the current is the same through all components.
Therefore, the current in the $5 \, \Omega$ resistor is $4 \, \text{A}$ .
The correct option is A.
$\boxed{4 \text{ A}}$
7. The potential difference across the 5 Ω resistor is ________.
From the previous question (Question 6), we found that the current $I$ flowing through the series circuit is $4 \, \text{A}$ .
To find the potential difference $V_1$ across the $5 \, \Omega$ resistor ( $R_1$ ), we use Ohm's Law
$V_1 = I \times R_1$
Substitute the values
$I = 4 \, \text{A}$ and $R_1 = 5 \, \Omega$ .
$V_1 = 4 \, \text{A} \times 5 \, \Omega$ $V_1 = 20 \, \text{V}$
The correct option is A.
$\boxed{20 \text{ V}}$
8. A current of 9 A flows when a 120 V battery is connected across a 12 Ω resistor. The battery has an internal resistance of ________.
For a circuit with a battery having an electromotive force (EMF) $\mathcal{E}$ and internal resistance $r$ , connected to an external resistor $R$ , the current $I$ flowing in the circuit is given by
$I = \frac{\mathcal{E}}{R + r}$
We are given
$I = 9 \, \text{A}$ , $\mathcal{E} = 120 \, \text{V}$ , and $R = 12 \, \Omega$ . We need to find $r$ .
Rearrange the formula to solve for $r$ :
$I(R + r) = \mathcal{E}$ $R + r = \frac{\mathcal{E}}{I}$ $r = \frac{\mathcal{E}}{I} - R$
Substitute the values
$r = \frac{120 \, \text{V}}{9 \, \text{A}} - 12 \, \Omega$ $r = 13.333... \, \Omega - 12 \, \Omega$ $r = 1.333... \, \Omega$
Rounding to one decimal place, $r \approx 1.3 \, \Omega$ .
The correct option is A.
$\boxed{1.3 \, \Omega}$
9. A 10 Ω resistor and a 5 Ω resistor are connected in parallel. Their equivalent resistance is ________.
For two resistors connected in parallel, the equivalent resistance $R_{eq}$ is given by the formula
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$
Substitute the given values
$R_1 = 10 \, \Omega$ and $R_2 = 5 \, \Omega$ .
$\frac{1}{R_{eq}} = \frac{1}{10 \, \Omega} + \frac{1}{5 \, \Omega}$
Find a common denominator and add the fractions
$\frac{1}{R_{eq}} = \frac{1}{10 \, \Omega} + \frac{2}{10 \, \Omega}$ $\frac{1}{R_{eq}} = \frac{3}{10 \, \Omega}$
Invert the result to find $R_{eq}$
$R_{eq} = \frac{10}{3} \, \Omega$ $R_{eq} \approx 3.33 \, \Omega$
The correct option is B.
$\boxed{3.3 \, \Omega}$

The resistance of a 120 V light bulb that draws a current of 1.25 A is .

Answer

Step-by-step solution

According to Ohm's Law, the resistance $R$ is given by

Substitute the given values

The resistance $R$ of a wire is given by the formula

The cross-sectional area $A$ of a circular wire is given by $A = \pi r^2$ , where $r$ is the radius.

Substitute the area into the resistance formula

For the second wire, the diameter is $d' = 2d$ .

From Step 3, we know that $R = \frac{4 \rho L}{\pi d^2}$ .

The resistors $5 \, \Omega$ and $10 \, \Omega$ are connected in series.

A potential difference $V = 60 \, \text{V}$ is applied across the series combination.

In a series circuit, the current is the same through all components.

From the previous question (Question 6), we found that the current $I$ flowing through the series circuit is $4 \, \text{A}$ .

To find the potential difference $V_1$ across the $5 \, \Omega$ resistor ( $R_1$ ), we use Ohm's Law

Substitute the values

For a circuit with a battery having an electromotive force (EMF) $\mathcal{E}$ and internal resistance $r$ , connected to an external resistor $R$ , the current $I$ flowing in the circuit is given by

We are given

Substitute the values

For two resistors connected in parallel, the equivalent resistance $R_{eq}$ is given by the formula

Substitute the given values

Find a common denominator and add the fractions

Invert the result to find $R_{eq}$

Try one yourself

Still stuck on this one?

More Physics Questions

The resistance of a 120 V light bulb that draws a current of 1.25 A is .

Answer

Step-by-step solution

According to Ohm's Law, the resistance RRR is given by

Substitute the given values

The resistance RRR of a wire is given by the formula

The cross-sectional area AAA of a circular wire is given by A=πr2A = \pi r^2A=πr2, where rrr is the radius.

Substitute the area into the resistance formula

For the second wire, the diameter is d′=2dd' = 2dd′=2d.

From Step 3, we know that R=4ρLπd2R = \frac{4 \rho L}{\pi d^2}R=πd24ρL​.

The resistors 5 Ω5 \, \Omega5Ω and 10 Ω10 \, \Omega10Ω are connected in series.

A potential difference V=60 VV = 60 \, \text{V}V=60V is applied across the series combination.

In a series circuit, the current is the same through all components.

From the previous question (Question 6), we found that the current III flowing through the series circuit is 4 A4 \, \text{A}4A.

To find the potential difference V1V_1V1​ across the 5 Ω5 \, \Omega5Ω resistor (R1R_1R1​), we use Ohm's Law

Substitute the values

For a circuit with a battery having an electromotive force (EMF) E\mathcal{E}E and internal resistance rrr, connected to an external resistor RRR, the current III flowing in the circuit is given by

We are given

Substitute the values

For two resistors connected in parallel, the equivalent resistance ReqR_{eq}Req​ is given by the formula

Substitute the given values

Find a common denominator and add the fractions

Invert the result to find ReqR_{eq}Req​

Try one yourself

Still stuck on this one?

More Physics Questions

According to Ohm's Law, the resistance $R$ is given by

The resistance $R$ of a wire is given by the formula

The cross-sectional area $A$ of a circular wire is given by $A = \pi r^2$ , where $r$ is the radius.

For the second wire, the diameter is $d' = 2d$ .

From Step 3, we know that $R = \frac{4 \rho L}{\pi d^2}$ .

The resistors $5 \, \Omega$ and $10 \, \Omega$ are connected in series.

A potential difference $V = 60 \, \text{V}$ is applied across the series combination.

From the previous question (Question 6), we found that the current $I$ flowing through the series circuit is $4 \, \text{A}$ .

To find the potential difference $V_1$ across the $5 \, \Omega$ resistor ( $R_1$ ), we use Ohm's Law

For a circuit with a battery having an electromotive force (EMF) $\mathcal{E}$ and internal resistance $r$ , connected to an external resistor $R$ , the current $I$ flowing in the circuit is given by

For two resistors connected in parallel, the equivalent resistance $R_{eq}$ is given by the formula

Invert the result to find $R_{eq}$