Chapter 23, Problem 77A

(a)

To determine

the current in the circuit when four lamps are turned on.

Answer to Problem 77A

The total current passes through the circuit is $i=2\text{A}$ .

Explanation of Solution

Given:

Power of lamps is 60 W.

Resistance of each lamp is $240\Omega$ .

Resistance of the heater is $10\Omega$ .

The potential difference across the circuit is 120 V.

Formula used:

Equivalent resistance of a circuit containing resistor in parallel is found out using the following formula:

  $\frac{1}{R_{\text{equi}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\mathrm{...}$  ......(1)

Here, $R_1,R_2,R_3$ are the resistances of the resistors.

Current in a circuit is equal to the voltage across the circuit divided by the equivalent resistance of the circuit that is,

  $i=\frac{V}{R_{\text{equi}}}$  ......(2)

Calculation:

Since all the four lamps are having same resistance this can be written as follows

  $\begin{array}{l}\frac{1}{R_{\text{equi}}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}\\ \frac{1}{R_{\text{equi}}}=\frac{4}{R}\end{array}$

Here, $R$ is the resistance of each lamp.

This gives,

  ${\mathrm{R}}_{\text{equi}}=\frac{R}{4}$  ......(3)

Substitute $R$ in equation (3)

  $\begin{array}{l}{\mathrm{R}}_{\text{equi}}=\frac{240}{4}\\ {\mathrm{R}}_{\text{equi}}=60\Omega \end{array}$

The current in the circuit is equal to the voltage across the circuit divided by the equivalent resistance of the circuit.

Substitute the values of $V$ and $R_{{}_{equi}}$ in equation (2)

  $\begin{array}{l}i=\frac{120\text{V}}{60\Omega }\\ \boxed{i=2\text{A}}\end{array}$

Conclusion:

Hence, the total current passes through the circuit is $i=2\text{A}$ .

(b)

To determine

the current in the circuit when all the lamps are turned on.

Answer to Problem 77A

The total current passes through the circuit is $i=3\text{A}$ .

Explanation of Solution

Given:

Power of lamps is 60 W.

Resistance of each lamp is $240\Omega$ .

Resistance of the heater is $10\Omega$ .

The potential difference across the circuit is 120 V.

Formula used:

Equivalent resistance of a circuit containing resistor in parallel is found out using the following formula:

  $\frac{1}{R_{\text{equi}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\mathrm{...}$  ......(1)

Here, $R_1,R_2,R_3$ are the resistances of the resistors.

Current in a circuit is equal to the voltage across the circuit divided by the equivalent resistance of the circuit that is,

  $i=\frac{V}{R_{\text{equi}}}$  ......(2)

Calculation:

The equivalent resistance of six equivalent lamps connected in parallel is equal to the resistance of each lamp divided by 6.

  ${\mathrm{R}}_{\text{equi}}=\frac{R}{6}$  ......(3)

Substitute $R$ in equation (3)

  $\begin{array}{l}{\mathrm{R}}_{\text{equi}}=\frac{240}{6}\\ {\mathrm{R}}_{\text{equi}}=40\Omega \end{array}$

The current in the circuit is equal to the voltage across the circuit divided by the equivalent resistance of the circuit.

Substitute the values of $V$ and $R_{{}_{equi}}$ in equation (2)

  $\begin{array}{l}i=\frac{120\text{V}}{40\Omega }\\ \boxed{i=3\text{A}}\end{array}$

Conclusion:

Hence, the total current passes through the circuit is $i=3\text{A}$ .

(c)

To determine

the current in the circuit when six lamps and the heater are operating.

Answer to Problem 77A

The total current passes through the circuit is $i=15\text{A}$ .

Explanation of Solution

Given:

Power of lamps is 60 W.

Resistance of each lamp is $240\Omega$ .

Resistance of the heater is $10\Omega$ .

The potential difference across the circuit is 120 V.

Formula used:

Equivalent resistance of a circuit containing resistor in parallel is found out using the following formula:

  $\frac{1}{R_{\text{equi}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\mathrm{...}$  ......(1)

Here, $R_1,R_2,R_3$ are the resistances of the resistors.

Current in a circuit is equal to the voltage across the circuit divided by the equivalent resistance of the circuit that is,

  $i=\frac{V}{R_{\text{equi}}}$  ......(2)

Calculation:

When six lamps and the water heater are connected in parallel , the equivalent resitance of the circuit is given by the following formula,

  $\frac{1}{{\mathrm{R}}_{\text{equi}}}=\frac{6}{R}+\frac{1}{{\mathrm{R}}_{\text{heater}}}$  ......(3)

Here, $R_{\text{heater}}$ is the resistance of the water heater.

Substitute the value of $R$ and $R_{\text{heater}}$ in equation (3)

  $\begin{array}{c}\frac{1}{{\mathrm{R}}_{\text{equi}}}=\frac{6}{240\Omega }+\frac{1}{10\Omega }\\ =\frac{6+24}{240\Omega }\\ =\frac{30}{240\Omega }\end{array}$

Take reciprocal of the above equation,

  $\begin{array}{l}{R}_{{}_{equi}}=\frac{240\Omega }{30}\\ R_{{}_{equi}}=8.0\Omega \end{array}$

The current in the circuit is equal to the voltage across the circuit divided by the equivalent resistance of the circuit.

  $i=\frac{V}{R_{equi}}$  ......(4)

Substitute the values of $V$ and $R_{equi}$ in equation (2)

  $\begin{array}{l}i=\frac{120\text{V}}{8.0\Omega }\\ \boxed{i=15\text{A}}\end{array}$

Conclusion:

Hence, the total current passes through the circuit is $i=15\text{A}$ .

(d)

To determine

whether 12 A fuse will get melted or not when all the lamps and the heater are on.

Answer to Problem 77A

The fuse won’t get melted.

Explanation of Solution

Given:

Total current when all the six lamps and the heater are turned on is $i=15\text{A}$ .

Current of fuse in the circuit is 12 A.

Calculation:

Since the current of fuse is less than the total current, so the fuse won’t get melted over time.

Conclusion:

Hence, the fuse won’t get melted.