Chapter 22, Problem 56A

a.

To determine

Ammeter reading.

Answer to Problem 56A

  $I=0.5\text{A}$

Explanation of Solution

Given:

The given figure is,

  

Here, source voltage is $V=9\text{V}$ and $R=18\Omega$

Formula used:

From Ohm’s law,

  $V=IR$

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

Calculation:

On substituting the given values in above relation,

  $\begin{array}{l}9=I\times 18\\ \Rightarrow I=\frac{9}{18}=0.5\text{A}\end{array}$

Conclusion:

Therefore, the ammeter reading is $I=0.5\text{A}$ .

b.

To determine

Voltmeter reading.

Answer to Problem 56A

9 V

Explanation of Solution

Given:

The given figure is,

  

Here, $V=9\text{V}$ and $R=18\Omega$

Calculation:

Consider the given circuit. It is observed that only 1 resistor of value $18\Omega$ is connected in the circuit. So, the voltage across resistor is same as the source voltage and equal to,

  $V_{18\Omega }=V_S=9\text{V}$

The voltmeter is connected across the resistor. So, the voltmeter reading will be

  $V=9\text{V}$

Conclusion:

The voltmeter reading is 9 V.

c.

To determine

Power delivered to the resistor.

Answer to Problem 56A

  $P=4.5\text{W}$

Explanation of Solution

Given:

The given figure is,

  

Here, $V=9\text{V}$ and $R=18\Omega$

Formula used:

The power is given as,

  $P=VI$

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

Calculation:

From the previous subpart, the current through the resistor is

  $I=0.5\text{A}$

On substituting the values,the power delivered is

  $\begin{array}{c}P=9\times 0.5\\ =4.5\text{W}\end{array}$

Conclusion:

Therefore, the power delivered to resistor is $P=4.5\text{W}$ .

d.

To determine

Energy delivered to resistor per hour.

Answer to Problem 56A

  $E=162\text{kJ}$

Explanation of Solution

Given:

The given figure is,

  

Here, $V=9\text{V}$ and $R=18\Omega$

Time, $t=1\text{hour}=3600\text{s}$

Formula used:

The energy power relation is,

  $E=Pt$

Where, $E$ is energy, $P$ is power and $t$ is time.

Calculation:

As, $P=4.5\text{W}$ from previous subpart.

So, on substituting the values the energy delivered to resistor is

  $\begin{array}{c}E=4.5\times 3600\\ =162\text{kJ}\end{array}$

Conclusion:

Therefore, the energy delivered to resistor is $E=162\text{kJ}$ .