Chapter 22, Problem 67A

a.

To determine

To identify: Whether the lamp obeys Ohm’s law or not.

Answer to Problem 67A

No, the lamp will not obey Ohm’s law.

Explanation of Solution

Given:

When the lamp is connected to voltage $V_1=6\text{V}$ , the current $I_1=66\text{mA}$ and

When it is connected to voltage $V_2=9\text{V}$ , the current $I_2=75\text{mA}$ .

Formula used:

Ohm’s law is given as,

  $V=IR$

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

Calculation:

From above relation, $R=\frac{V}{I}$

Now, $R_1=\frac{6}{66\times {10}^{-3}}=90.9\Omega$

And $R_2=\frac{9}{75\times {10}^{-3}}=120\Omega$

Since, $R_1\ne R_2$ it means the lamp does not follow Ohm’s law.

Conclusion:

As in Ohm’s law $R$ must be constant but $R_1\ne R_2$ it means the lamp does not follow Ohm’s law.

b.

To determine

Power dissipated by lamp.

Answer to Problem 67A

  $P_1=0.396\text{W}$

Explanation of Solution

Given:

When the lamp is connected to voltage $V_1=6\text{V}$ , the current $I_1=66\text{mA}$ and

When it is connected to voltage $V_2=9\text{V}$ , the current $I_2=75\text{mA}$

Formula used:

The power is given as,

  $P=VI$

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

Calculation:

When lamp is connected to $V_1=6\text{V}$ battery the current is $I_1=66\text{mA}$

So, the power dissipated is

  $\begin{array}{c}{P}_1=6\times 66\times {10}^{-3}\\ =0.396\text{W}\end{array}$

Conclusion:

Therefore, the power dissipated in lamp when connected to 6 V battery is $P_1=0.396\text{W}$ .