Chapter 22.1, Problem 18PP

a.

To determine

Potential difference across the lamp.

Answer to Problem 18PP

  $V=62.5\text{V}$

Explanation of Solution

Given:

Power of lamp, $P=75\text{W}$

Supply voltage, $V_S=125\text{V}$

Original value of current, $I=0.6\text{A}$

Resistance of lamp, $R_{\text{lamp}}=208.33\Omega$

Formula used:

The Ohm’s law is given as, $V=IR$

Where, $V$ is the voltage, $I$ is the current and $R$ is the resistance.

Calculation:

As a resistor is added to reduce the current to half of the original value.

So, the new value of current is

  $I_{\text{new}}=\frac{0.6}{2}=0.3\text{A}$

Now, the voltage across the lamp is

  $\begin{array}{c}V=IR\\ =0.3\times 208.33\\ =62.5\text{V}\end{array}$

Conclusion:

Therefore, the potential difference across the lamp is $V=62.5\text{V}$ .

b.

To determine

Value of resistance added to the circuit.

Answer to Problem 18PP

  $R_{\text{res}}=208.34\Omega$

Explanation of Solution

Given:

Power of lamp, $P=75\text{W}$

Supply voltage, $V_S=125\text{V}$

Original value of current, $I=0.6\text{A}$

Resistance of lamp, $R_{\text{lamp}}=208.33\Omega$

Formula used:

The Ohm’s law is given as, $V=IR$

Where, $V$ is the voltage, $I$ is the current and $R$ is the resistance.

Calculation:

The total resistance of the circuit is,

  $\begin{array}{c}{R}_{\text{total}}=\frac{V_S}{I_{\text{new}}}=\frac{125}{0.3}\\ =416.67\Omega \end{array}$

So, resistance of lamp is

  $\begin{array}{c}{R}_{\text{res}}=R_{\text{total}}-R_{\text{lamp}}\\ =416.7\Omega -208.33\Omega \\ =208.34\Omega \end{array}$

Conclusion:

Therefore, the resistance of the resistor is $R_{\text{res}}=208.34\Omega$ .

c.

To determine

Rate of electrical energy transformed into radiant and thermal energy by lamp.

Answer to Problem 18PP

  $P=18.75\text{W}$

Explanation of Solution

Given:

Power of lamp, $P=75\text{W}$

Supply voltage, $V_S=125\text{V}$

Original value of current, $I=0.6\text{A}$

Resistance of lamp, $R_{\text{lamp}}=208.33\Omega$

Formula used:

The rate of transformation of energy is power which is given as, $P=VI$

Where, $V$ is the voltage across the lamp and $I$ is current through the lamp.

Calculation:

As, from previous subpart

The potential difference across the lamp is $V=62.5\text{V}$

And current through the lamp is $I_{\text{new}}=0.3\text{A}$

So, power is

  $\begin{array}{l}P=62.5\times 0.3\\ =18.75\text{W}\end{array}$

Conclusion:

Therefore, the rate of transformation of energy i.e. power is $P=18.75\text{W}$ .