Chapter 25.1, Problem 5PP

(a)

To determine

To calculate:Effective potential difference.

Answer to Problem 5PP

Effective potential differenceis $120\text{V}$ .

Explanation of Solution

Given:

Maximum potential difference, $V_{\max }=170\text{V}$

Formula used:

The formula to calculate effective voltage is,

  $V_{eff}=0.707\left(V_{\max }\right)$

Where,

  $V_{eff}=$ Effective potential difference

  $V_{\max }=$ Maximum potential difference

Calculation:

Effective potential difference can be calculated by the formula,

  $V_{eff}=0.707\left(V_{\max }\right)=0.707\left(170\text{V}\right)=120\text{V}$

Conclusion:

Effective potential differenceis $120\text{V}$ .

(b)

To determine

To calculate:Effective current through the lamp.

Answer to Problem 5PP

Effective current of the lamp is $\text{0}\text{.49}\text{A}$

Explanation of Solution

Given:

Maximum current, $I_{\max }=0.70\text{A}$

Formula used:

Effective current is calculated by the formula $I_{eff}=0.707\left(I_{\max }\right)$

Where,

  $I_{eff}=$ Effective current

  $I_{\max }=$ Maximum current

Calculation:

The formula for effective current used is,

  

Substituting the values in above formula,

  $\begin{array}{l}{I}_{eff}=0.707\left(I_{\max }\right)\\ =\left(0.707\right)\left(0.70\text{A}\right)\\ =0.49\text{A}\end{array}$

Conclusion:

Effective current of the lamp is $0.49\text{A}$ .

(c)

To determine

To calculate:Resistance of the lamp.

Answer to Problem 5PP

Resistance of the lamp is $2.4\times {10}^2\Omega$

Explanation of Solution

Given:

For a lamp,

Effective voltage, $V_{eff}=120\text{V}$

Effective current, ${\text{I}}_{eff}=0.49\text{A}$

Formula used:

Resistance is calculated by $R=\frac{V_{eff}}{I_{eff}}$

Where,

  $V_{eff}=$ Effective potential difference

  $I_{eff}=$ Effective current

Calculation:

Resistance of the lamp can be calculated by the formula,

  $R=\frac{V_{eff}}{I_{eff}}=\frac{\frac{V_{\max }}{\sqrt{2}}}{\frac{I_{\max }}{\sqrt{2}}}=\frac{V_{\max }}{I_{\max }}=\frac{170\text{V}}{0.70\text{A}}=2.4\times {10}^2\Omega$

Conclusion:

Resistance of the lamp is $2.4\times {10}^2\Omega$ .