To calculate the time it takes to charge a battery, we use the following formula:

$$\text{Time} = \frac{\text{Battery Capacity (Ah)}}{\text{Charging Current (A)}}$$

Given:

Battery Capacity = 200 Ah; Charging Current = 25 A

Plugging in the values:

$$\text{Time} = \frac{200\ \text{Ah}}{25\ \text{A}} = 8\ \text{hours}$$

Therefore, it will take 8 hours to charge the battery.

Given:

• Voltage (V) = 12.0 V
• Current (I) = 2.0 A

We need to find the resistance (R). We can use Ohm’s Law, which states:

$$V = I \times R$$

We need to rearrange the formula to solve for R.

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

Now, we can plug in the given values:

$$R = \frac{12.0\ \text{V}}{2.0\ \text{A}} \Rightarrow R = 6.0\ \Omega$$

So, the resistance offered by the electrical appliance is 6.0 Ω.

We need to determine how the current (I) changes.

Ohm’s Law states:

$$V = I \times R$$

We can rearrange this to solve for the current I.

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

Let’s denote the initial potential difference as V₁ and the final potential difference as V₂. Similarly, let I₁ be the initial current and I₂ be the final current.

The resistance R remains constant.

We are given that the final potential difference is one-fourth of the initial potential difference:

$$V_2 = \frac{1}{4}V_1$$

The initial current I₁ is:

$$I_1 = \frac{V_1}{R}$$

The final current I₂ is:

$$I_2 = \frac{V_2}{R}$$

Substitute $$V_2 = \frac{1}{4}V_1$$ into the equation for I₂.

$$I_2 = \frac{\frac{1}{4}V_1}{R} = \frac{1}{4}\frac{V_1}{R}$$

Since $$I_1 = \frac{V_1}{R}$$, we can write: $$I_2 = \frac{1}{4}I_1$$

This means the final current I₂ is one-fourth of the initial current I₁.

Therefore, the current through the component will decrease to one-fourth of its initial value.

Given:

• Current in amperes (A) = I = 50 A
• Resistance in ohms (Ω) = R = 100 Ω
• Voltage in volts (V) = ?

Substitute the given values into Ohm’s Law:

$$V = I \times R \quad \Rightarrow \quad V = 50\ \text{A} \times 100\ \Omega \quad \Rightarrow \quad V = 5000\ \text{V}$$

Therefore, the voltage necessary to pass a current of 50 amperes through a resistance of 100 Ω is 5000 V.

$$V = I \times R$$

First, we need to find the resistance of the resistor using the initial values of voltage and current:

$$V = I \times R \quad \Rightarrow \quad R = \frac{V}{I}$$

Given $$V_1 = 6.0\ \text{V}$$ and $$I_1 = 0.5\ \text{A}$$, we can calculate the resistance:

$$R = \frac{6.0\ \text{V}}{0.5\ \text{A}} = 12\ \Omega$$

Now that we have the resistance, we can find the new current ($$I_2$$) when the voltage is $$V_2 = 9.0\ \text{V}$$.

Using Ohm’s Law again:

$$I_2 = \frac{V_2}{R} \quad \Rightarrow \quad I_2 = \frac{9.0\ \text{V}}{12\ \Omega} = 0.75\ \text{A}$$

So, the current that will pass through the resistor when the voltage across it is 9.0 V is 0.75 A.

Given:

• Resistance = $$R = 0.085\ \Omega$$
• Length of the wire = $$L = 2\ \text{m}$$
• Cross-sectional area of the wire = $$A = 0.4\ \text{mm}^2 = 0.4 \times 10^{-6}\ \text{m}^2$$
• Resistivity = $$\rho = ?$$

We need to solve for $$\rho$$.

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

Plugging in the values:

$$\rho = \frac{0.085\ \Omega \times 0.4 \times 10^{-6}\ \text{m}^2}{2\ \text{m}} = \frac{0.085 \times 0.4 \times 10^{-6}}{2}\ \Omega \cdot \text{m}$$

$$\rho = \frac{0.034 \times 10^{-6}}{2}\ \Omega \cdot \text{m} = 0.017 \times 10^{-6}\ \Omega \cdot \text{m}$$

$$\rho = 1.7 \times 10^{-8}\ \Omega \cdot \text{m}$$

So, the resistivity of the wire is 1.7 × 10⁻⁸ Ω · m.

(a)

If the wire is stretched to double its length, what will be the new resistivity?

$$\rho = 1.7 \times 10^{-8}\ \Omega\text{m}$$

Resistivity is a material property and does not change when the length of the wire is changed (assuming the material and temperature remain constant).

Stretching the wire changes its dimensions (length and cross sectional area), but not its inherent ability to resist the flow of electric current.

Therefore, the new resistivity will be the same as the initial resistivity.

$$\rho_{\text{new}} = 1.7 \times 10^{-8}\ \Omega \cdot \text{m}$$

(b)

If its temperature is increased by 20 K, find the new resistivity.

Given:

• Resistivity at the reference temperature $$T_o = \rho_o = 1.7 \times 10^{-8}\ \Omega \cdot \text{m}$$
• Temperature coefficient of resistivity, $$\alpha = 3.9 \times 10^{-3}\ \text{K}^{-1}$$
• Change in temperature = $$T - T_o = 20\ \text{K}$$
• Resistivity at temperature $$T = \rho_T = ?$$

Plugging in the values:

$$\rho_T = \rho_o [1 + \alpha (T - T_o)]$$

$$\rho_T = 1.7 \times 10^{-8}[1 + (3.9 \times 10^{-3})(20)]$$

$$\rho_T = 1.7 \times 10^{-8}[1 + 0.078]$$

$$\rho_T = 1.7 \times 10^{-8}[1.078]$$

$$\rho_T = 1.8326 \times 10^{-8}\ \Omega \cdot \text{m}$$

Rounding to two significant figures:

$$\rho_T \approx 1.83 \times 10^{-8}\ \Omega \cdot \text{m}$$

So, the new resistivity after the temperature increase is 1.83 × 10⁻⁸ Ω · m.

Given:

• Resistivity at the initial temperature = $$T_o = \rho_o = 1.7 \times 10^{-8}\ \Omega \cdot \text{m}$$
• Temperature coefficient of resistivity = $$\alpha = 0.004\ \text{K}^{-1}$$
• Change in temperature = $$T - T_o = 50\ \text{K}$$
• Resistivity at the new temperature = $$T = \rho_T = ?$$

Plug in the values:

$$\rho_T = \rho_o [1 + \alpha (T - T_o)]$$

$$\rho_T = 1.7 \times 10^{-8}[1 + 0.004 \times 50]$$

$$\rho_T = 1.7 \times 10^{-8}[1 + 0.2]$$

$$\rho_T = 1.7 \times 10^{-8}[1.2]$$

$$\rho_T = 2.04 \times 10^{-8}\ \Omega \cdot \text{m}$$

So, the new resistivity is 2.04 × 10⁻⁸ Ω · m.