Question

When potential difference = 5.6v and the current =0.04a what is the temperature

Answer 1

Explanation: The information you provided (potential difference of 5.6 volts and current of 0.04 amperes) does not seem to be directly related to determining temperature. In physics or electrical engineering, the relationship between potential difference (voltage), current, and temperature would depend on the specific context or system you are referring to.

If you are talking about a resistor in an electrical circuit, for example, you might need additional information such as the resistance of the resistor (R) to use Ohm's Law, which states:

V=I×R

V is the potential difference (voltage),

I is the current,

R is the resistance.

Answer 2

The calculated change in temperature
`(\(\Delta T\))` is approximately
`\(0.1786\,^\circ C\).`

Determining the temperature in a wire from the given potential difference (V = 5.6V) and current (I = 0.04A) involves understanding the relationship between electrical power, resistance, and temperature change.

The power dissipated in the wire can be calculated using the formula P = IV, where P is power, \(I\) is current, and \(V\) is potential difference.

`\[ P = (0.04\,A) * (5.6\,V) = 0.224\,W \]`

Next, we need to consider the power equation
`\(P = I^2R\)`, where \(R\) is resistance.

Rearranging for resistance,
`\(R = (P)/(I^2)\):`

`\[ R = (0.224\,W)/((0.04\,A)^2) = 140\,\Omega \]`

Now, the temperature change
`(\(\Delta T\))` can be determined using the relationship between power, resistance, and temperature change given by
`\(P = I^2R\Delta T\):`

`\[ \Delta T = (P)/(I^2R) = (0.224\,W)/((0.04\,A)^2 * 140\,\Omega) \]`

The given expression is for calculating the change in temperature
`(\(\Delta T\))` using the formula
`\(\Delta T = (P)/(I^2R)\)`, where (P) is the power, (I) is the current, and (R) is the resistance.

Substitute the provided values into the formula:

`\[ \Delta T = (0.224\,W)/((0.04\,A)^2 * 140\,\Omega) \]\\ \\\Delta T = (0.224\,W)/(0.000064\,A^2 * 140\,\Omega) \]`

Simplify the expression:

`\[ \Delta T = (0.224\,W)/(0.00896\,A^2 * 140\,\Omega) \]\\\\\Delta T \approx (0.224\,W)/(1.2544\,A^2) \]\\\\\Delta T \approx 0.1786\, ^\circ C \]`

Therefore, The calculated change in temperature
`(\(\Delta T\))` is approximately
`\(0.1786\,^\circ C\).`