Chapter 25.1, Problem 6PP

To determine

To calculate:Maximum potential difference and maximum current of the lamp.

Answer to Problem 6PP

Maximum potential difference is $165\text{V}$ and maximum current is $7.8\text{A}$ .

Explanation of Solution

Given:

Effective voltage, $V_{eff}=117\text{V}$

Effective current, ${\text{I}}_{eff}=5.5\text{A}$

Formula Used:

Formula to calculate maximum voltage,

  $V_{\max }=\frac{V_{eff}}{0.707}$

Formula to calculate maximum current $I_{\max }=\frac{I_{eff}}{0.707}$

Where,

  $V_{eff}=$ Effective potential difference (voltage)

  $V_{\max }=$ Maximum voltage

  $I_{eff}=$ Effective current

  $I_{\max }=$ Maximum current

Calculation:

Maximum potential difference can be calculated by the formula,

  $V_{\max }=\frac{V_{eff}}{0.707}$

Substituting the values in above formula,

  $=\frac{117\text{V}}{0.707}=165\text{V}$

Maximum current can be calculated by the formula,

  $I_{\max }=\frac{I_{eff}}{0.707}$

Substituting the values in above formula,

  $=\frac{5.5\text{A}}{0.707}=7.8\text{A}$

Conclusion:

Maximum potential difference is $165\text{V}$ .

Maximum current is $7.8\text{A}$ .