Chapter 23, Problem 51A

a.

To determine

Equivalent resistance of the circuit.

Answer to Problem 51A

Equivalent resistance of circuit is, $R=26.5\Omega$ .

Explanation of Solution

Given:

Resistance of first lamp, $R_1=22\Omega$

Resistance of second lamp, $R_2=4.5\Omega$

Potential difference, $V=45\text{ V}$

Formula used:

For series circuit equivalent resistance is given by,

  $R=R_1+R_2+R_3+\mathrm{..........}{R}_n$

Where, $R$ is equivalent resistance and $R_1,R_2,R_3\mathrm{..........}{R}_n$ are individual resistances.

Calculation:

Now, substituting the value of $R_1,R_2$ and solve.

  $\begin{array}{c}R=22+4.5\\ =26.5\Omega \end{array}$

Conclusion:

Therefore, equivalent resistance of circuit is $26.5\Omega$ .

b.

To determine

Current flowing through the circuit.

Answer to Problem 51A

Current flowing through the circuit, $I=1.698\text{ A}$ .

Explanation of Solution

Given:

Resistance of first lamp, $R_1=22\Omega$

Resistance of second lamp, $R_2=4.5\Omega$

Potential difference, $V=45\text{ V}$

Formula used:

According to ohm’s law,

  $V=IR$

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

Calculation:

From part (a) $R=26.5\Omega$

Now, substituting the value of V and R and solve.

  $\begin{array}{c}45=I\times 26.5\\ I=\frac{45}{26.5}\\ =1.698\text{ A}\end{array}$

Conclusion:

Therefore, current flowing through circuit is 1.698 A.

c.

To determine

Potential difference across each lamp.

Answer to Problem 51A

Potential difference across first lamp, $V_1=37.36\text{ V}$

Potential difference across second lamp, $V_2=7.64\text{ V}$

Explanation of Solution

Given:

Resistance of first lamp, $R_1=22\Omega$

Resistance of second lamp, $R_2=4.5\Omega$

Potential difference, $V=45\text{ V}$

Formula used:

According to ohm’s law,

  $V=IR$

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

Calculation:

From part (b) $I=1.698\text{ A}$

Potential difference across first lamp.

  $\begin{array}{c}{V}_1=I{R}_1\\ =1.698\times 22\\ =37.36\text{ V}\end{array}$

Potential difference across second lamp.

  $\begin{array}{c}{V}_2=I{R}_2\\ =1.698\times 4.5\\ =7.64\text{ V}\end{array}$

Conclusion:

Therefore, potential difference across first lamp is 37.36 V and potential difference across second lamp is 7.64 V.

d.

To determine

Power used in each lamp.

Answer to Problem 51A

Power used by first lamp, $P_1=63.43\text{ W}$

Power used by second lamp, $P_2=12.97\text{ W}$

Explanation of Solution

Given:

Resistance of first lamp, $R_1=22\Omega$

Resistance of second lamp, $R_2=4.5\Omega$

Potential difference, $V=45\text{ V}$

Formula used:

Power is given by,

  $P=VI$

Where, $I$ is current and V is potential difference.

Calculation:

From part (b) $I=1.698\text{ A}$

From part (c) $V_1=37.36\text{ V, }{V}_2=7.64\text{ V}$

Power used by first lamp.

  $\begin{array}{c}{P}_1=V_{1}I\\ =37.36\times 1.698\\ =63.43\text{ W}\end{array}$

Power used by second lamp.

  $\begin{array}{c}{P}_2=V_{2}I\\ =7.64\times 1.698\\ =12.97\text{ W}\end{array}$

Conclusion:

Therefore, power used by first lamp is 63.43 W and power used by second lamp is 12.97 W.