Chapter 23.1, Problem 7PP

(a)

To determine

the current in the circuit.

Answer to Problem 7PP

The current in the circuit is 66.7 mA.

Explanation of Solution

Given:

  $R_1=255\Omega$ .

  $R_2=290\Omega$

  $\Delta V_1=17\text{V}$

The circuit diagram is

  

Formula:

The expression for potential is given by

  $V=IR$  ......(1)

Here, $V$ , $I$ and $R$ are potential difference, current and resistance.

Calculation:

Voltage drop $V_{\text{A}}$ against the resistor $R_{\text{A}}$ is

  $V_{\text{A}}=I{R}_{\text{A}}$  ......(2)

Solve for $I$ ,

Substitute the values in equation (2)

  $\begin{array}{l}I=\frac{V_{\text{A}}}{R_{\text{A}}}\\ I=\frac{17.0\text{V}}{255\Omega }\\ I=66.7\text{mA}\end{array}$

Thus, the current in the circuit is 66.7 mA.

Conclusion:

Hence, the required voltage drop across the 22 $\Omega$ resistor is 11 V.

(b)

To determine

the potential difference across the battery.

Answer to Problem 7PP

The required voltage drop across the voltage is 36.5 V.

Explanation of Solution

Given:

Reading of ammeter is $I=0.50\text{A}$ .

Resistance is $R_A=22\Omega$ .

Resistance is $R_B=22\Omega$ .

Formula:

The expression for potential is given by

  $V=IR$  ......(1)

Here, $V$ , $I$ and $R$ are potential difference, current and resistance.

Calculation:

The equivalent resistance is the sum of the resistance that is,

  $R=R_{\text{A}}+R_{\text{B}}$  ......(2)

Substitute the values,

  $\begin{array}{l}R=255\Omega +292\Omega \\ R=547\Omega \end{array}$

Now, substitute in equation (1)

  $\begin{array}{l}V=\left(66.7\text{mA}\right)\left(547\Omega \right)\\ \boxed{V=36.5\text{V}}\end{array}$

Conclusion:

Hence, the required voltage drop across the voltage is 36.5 V.

(c)

To determine

the total power used in the circuit and power used in each resistor.

Answer to Problem 7PP

Power dissipation is $\boxed{P=2.43\text{W}}$ and individual power dissipation are 1.13 W and 1.30 W respectively.

Explanation of Solution

Given:

Reading of ammeter is $I=0.50\text{A}$ .

Resistance is $R_A=22\Omega$ .

Resistance is $R_B=22\Omega$ .

Voltage drop across the 15 $\Omega$ resistor is 7.5 V.

Voltage drop across the 22 $\Omega$ resistor is 11 V.

Formula:

Power is given by the expression

  $P=IV$  ......(1)

Here, $I$ is current and $V$ is voltage drop.

Calculation:

From path (a) and (b), $I=66.7\text{mA}$ and V=36.5 V

Substitute the values in equation (1)

  $\begin{array}{l}P=\left(66.7\text{mA}\right)\left(\text{36}.\text{5 V}\right)\\ \boxed{P=2.43\text{W}}\end{array}$

Power dissipation across the resistor $R_{\text{A}}$ is

  $P=I{V}_{\text{A}}$  ......(2)

Substitute the values in equation (2)

  $\begin{array}{l}{P}_{\text{A}}=\left(66.7\text{mA}\right)\left(17.0\text{V}\right)\\ \boxed{P_{\text{A}}=1.13\text{W}}\end{array}$

Similarly,

Power dissipation across the resistor $R_{\text{B}}$ is

  $P=I{V}_{\text{B}}$  ......(3)

The voltage drop $V_{\text{s}}$ can be found by subtracting $V_{\text{A}}$ from the battery voltage,

  $\begin{array}{l}{V}_{\text{B}}=\left(36.5-17.0\right)\text{V}\\ V_{\text{B}}=19.5\text{V}\end{array}$

Substitute the values in equation (3)

  $\begin{array}{l}{P}_{\text{B}}=\left(66.7\text{mA}\right)\left(19.5\text{V}\right)\\ \boxed{P_{\text{B}}=1.30\text{W}}\end{array}$

Conclusion:

Hence, the required power dissipation is $\boxed{P=2.43\text{W}}$ and individual power dissipation are 1.13 W and 1.30 W respectively.

(d)

To determine

whether sum of the power used in each resistor in the circuit equal to the total power used in the circuit.

Answer to Problem 7PP

The rate at which energy is converted or power is dissipated will equal to the sum of parts.

Explanation of Solution

Given:

Reading of ammeter is $I=0.50\text{A}$ .

Resistance is $R_A=22\Omega$ .

Resistance is $R_B=22\Omega$ .

Voltage drop across the 15 $\Omega$ resistor is 7.5 V.

Voltage drop across the 22 $\Omega$ resistor is 11 V.

Total power dissipation is $P=2.43\text{W}$ .

Individual power dissipation are 1.13 W and 1.30 W.

Calculation:

The sum of the individual power dissipation is 1.13 W + 1.30 W = 2.43 W.

Thus, the sum of the individual power dissipation in the circuit equals the total power dissipation in the circuit and thereby the total power is conserved.

Conclusion:

Hence, the rate at which energy is converted or power is dissipated will equal to the sum of parts.