Detailed Answer

The potential difference is the work done per unit charge, hence we need to do $\frac{100J}{5C}$.

Detailed Answer

This is what Ohm’s law states, that these 2 quanitites must be directly proportional.

Detailed Answer

Since the first 2 resistors are in parallel, their equivalent resistance will be $\frac{1}{R} = \frac{1}{5} + \frac{1}{5}$ which gives $2.5\Omega$. Then the internal resistance of the battery is in series, so $2\Omega$ must be added, resulting in B.

Detailed Answer

For this, we need to use the equation $L = \frac{Q}{m}$, so we have $Q = 0.25 \cdot 3 \cdot 10^5$, which results in A.

Detailed Answer

We have the equation $R = \frac{p \cdot L}{A}$, and if we double the radius, the area will quadruple. Hence, we have $\frac{2}{4}$, so the resistance halves.

Detailed Answer

The temperature increases because of the current flowing, however this increase in temperature inturn also increases the resistance of the component, hence the overall current will decrease in the circuit.

Detailed Answer

It is important to understand that Ohms law does not describe a differential relationship between voltage and current, it only says it for particular points. (Imagine we have a graph of potential on the y axis and current on the x axis, and the graph is horizontal. That does not mean that the resistance of the component is 0 at those points, right?)

Detailed Answer

The electric potential difference $V$ is given by the equation:

$$V = \frac{W}{Q}$$

Where: $W = 10.0 \, \text{J}$, $Q = 2.0 \, \text{C}$

Substituting the values:

$$V = \frac{10.0}{2.0} = 5.0 \, \text{V}$$

Detailed Answer

According to Ohm's law:

$$V = IR$$

Rearranging for $R$:

$$R = \frac{V}{I}$$

Where: $V = 10.0 \, \text{V}$, $I = 0.50 \, \text{A}$

Substituting the values:

$$R = \frac{10.0}{0.50} = 20.0 \, \Omega$$

Detailed Answer

The resistivity $\rho$ is given by the formula:

$$R = \rho \frac{L}{A}$$

Rearranging for $\rho$:

$$\rho = \frac{R A}{L}$$

Where: $R = 10.0 \, \Omega$, $L = 2.0 \, \text{m}$, $A = 1.0 \times 10^{-6} \, \text{m}^2$.

Substitute the values:

$$\rho = \frac{(10.0)(1.0 \times 10^{-6})}{2.0} = 5.0 \times 10^{-5} \, \Omega \cdot \text{m}$$

Detailed Answer

An ohmic conductor follows Ohm's law, meaning the current is directly proportional to the potential difference and the resistance remains constant regardless of the applied voltage or current direction.

Detailed Answer

For resistors in parallel, the total resistance $R_{\text{total}}$ is given by:

$$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$

Substitute the values:

$$\frac{1}{R_{\text{total}}} = \frac{1}{2.0} + \frac{1}{4.0} + \frac{1}{6.0} = 0.5 + 0.25 + 0.167 = 0.917$$

$$R_{\text{total}} = \frac{1}{0.917} \approx 1.09 \, \Omega$$

Detailed Answer

For resistors in series, the total resistance $R_{\text{total}}$ is:

$$R_{\text{total}} = R_1 + R_2 + R_3 = 4.0 + 6.0 + 8.0 = 18.0 \, \Omega$$

The total power dissipated in the resistors is given by the formula:

$$P = \frac{V^2}{R_{\text{total}}}$$

where:

• $V = 24.0 \, \text{V}$
• $R_{\text{total}} = 18.0 \, \Omega$

Substitute the values:

$$P = \frac{(24.0)^2}{18.0} = \frac{576}{18.0} = 32.0 \, \text{W}$$

Detailed Answer

For resistors in parallel, the total resistance $R_{\text{total}}$ is given by:

$$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$$

Substitute the values:

$$\frac{1}{R_{\text{total}}} = \frac{1}{3.0} + \frac{1}{6.0} + \frac{1}{9.0} + \frac{1}{12.0}$$

$$\frac{1}{R_{\text{total}}} = 0.333 + 0.167 + 0.111 + 0.083 = 0.694$$

$$R_{\text{total}} = \frac{1}{0.694} \approx 1.44 \, \Omega$$

The total power dissipated is:

$$P = \frac{V^2}{R_{\text{total}}}$$

where:

• $V = 36.0 \, \text{V}$
• $R_{\text{total}} = 1.44 \, \Omega$

Substitute the values:

$$P = \frac{(36.0)^2}{1.44} = \frac{1296}{1.44} = 900.0 \, \text{W}$$