Chapter 23, Problem 49A

(a)

To determine

the resistance of the circuit.

Answer to Problem 49A

The total resistance of the circuit is $\boxed{R_{\text{total}}=37\Omega }$ .

Explanation of Solution

Given:

The ammeter 1 reads 0.20 A in the circuit is shown below figure:

  

  $R_1=22\Omega$

  $R_2=15\Omega$

Formula used:

Total resistance can be written as

  $R_{\text{total}}=R_1+R_2$  ......(1)

Calculation:

The two resistors in the circuits are connected in series. The total resistance of the circuit with two resistor connected in series is equal to the sum of the resistance of the two resistors.

Substitute the values of $R_1$ and $R_2$ in equation (1)

  $\begin{array}{l}{R}_{\text{total}}=\left(22+15\right)\Omega \\ \boxed{R_{\text{total}}=37\Omega }\end{array}$

Conclusion:

Hence, the total resistance of the circuit is $\boxed{R_{\text{total}}=37\Omega }$ .

(b)

To determine

the potential difference across the battery.

Answer to Problem 49A

The total voltage of the battery is 7.4 V.

Explanation of Solution

Given:

The ammeter 1 reads 0.20 A in the circuit is shown below figure:

  

  $R_1=22\Omega$

  $R_2=15\Omega$

  $R_{\text{total}}=37\Omega$

  $i=0.20\text{A}$

Formula used:

Total resistance can be written as

  $V=i{R}_{\text{total}}$  ......(1)

Calculation:

As all the three ammeters in the circuit are in series, all of them slow the same reading. Voltage of the battery is equal to the current in the circuit multiplied by the total resistance of the circuit.

Substitute the values of $i$ and $R_{\text{total}}$ in equation (1)

  $\begin{array}{l}V=\left(20\text{A}\right)\left(37\Omega \right)\\ \boxed{V=7.4\text{V}}\end{array}$

Conclusion:

Hence, the total voltage of the battery is 7.4 V.

(c)

To determine

the power delivered to the 22 $\Omega$ resistor..

Answer to Problem 49A

The total power delivered to the resistor is 0.88 W.

Explanation of Solution

Given:

The ammeter 1 reads 0.20 A in the circuit is shown below figure:

  

  $R_1=22\Omega$

  $R_2=15\Omega$

  $R_{\text{total}}=37\Omega$

  $i=0.20\text{A}$

Formula used:

The power delivered to a resistor is equal to the square of the current through it mulplited by the resistance of that resistor.

  $P=i^{2}R$  ......(1)

Calculation:

Substitute the values of $i$ and $R_1$

  $\begin{array}{l}P={\left(0.20\text{A}\right)}^2\left(22\Omega \right)\\ \boxed{P=0.88\text{W}}\end{array}$

Conclusion:

Hence, the total power delivered to the resistor is 0.88 W.

(d)

To determine

the power delivered by the battery.

Answer to Problem 49A

The total power delivered by the battery is 1.48 W.

Explanation of Solution

Given:

The ammeter 1 reads 0.20 A in the circuit is shown below figure:

  

  $R_1=22\Omega$

  $R_2=15\Omega$

  $R_{\text{total}}=37\Omega$

  $i=0.20\text{A}$

Formula used:

The power delivered to a resistor is equal to the square of the current through it mulplited by the resistance of that resistor.

  $P=iV$  ......(1)

Calculation:

Substitute the values of $i$ and $V$

  $\begin{array}{l}P={\left(7.4\text{V}\right)}^2\left(0.20\text{A}\right)\\ \boxed{P=1.48\text{W}}\end{array}$

Conclusion:

Hence, the total power delivered by the battery is 1.48 W.