Chapter 23, Problem 50A

(a)

To determine

The potential difference across the 22 $\Omega$ resistor.

Answer to Problem 50A

Voltage drop across the 22 $\Omega$ resistor is 11 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:

To find the voltage drop across $R_A$ , equation (1) becomes,

  $V_A=I{R}_A$  ......(2)

Substitute $0.50\text{A}$ for $I$ and $22\Omega$ for $R_A$ in equation (2)

  $\begin{array}{l}{V}_A=\left(0.50\text{A}\right)\times \left(22\Omega \right)\\ \boxed{V_A=11\text{V}}\end{array}$

Conclusion:

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

(b)

To determine

The potential difference across the 15 $\Omega$ resistor.

Answer to Problem 50A

Voltage drop across the 15 $\Omega$ resistor is 7.5 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:

To find the voltage drop across $R_B$ , equation (1) becomes,

  $V_B=I{R}_B$  ......(2)

Substitute $0.50\text{A}$ for $I$ and $15\Omega$ for $R_B$ in equation (2)

  $\begin{array}{l}{V}_A=\left(0.50\text{A}\right)\times \left(15\Omega \right)\\ \boxed{V_A=7.5\text{V}}\end{array}$

Conclusion:

Hence, the required voltage drop across the 15 $\Omega$ resistor is 7.5 V.

(c)

To determine

The potential difference across the battery.

Answer to Problem 50A

The potential difference across the voltage is 18.5 V.

Explanation of Solution

Given:

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

Resistance is $R_A=22\Omega$ .

Resistance is $R_B=22\Omega$ .

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

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

Formula:

The expression for potential is given by

  $V=IR$  ......(1)

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

Calculation:

The battery voltage is equal to the sum of the voltage drop across the individual resistor. The sum of the voltage drop across the two resistor is 11 V +7.5 V = 18.5 V.

Conclusion:

Hence, the required potential difference across the voltage is 18.5 V.