Question

Part A Consider a circuit containing five identical light bulbs and an ideal battery. Assume that the resistance of each light bulb remains constant. Rank the bulbs (A through E) based on their brightness. Rank from brightest to dimmest. To rank items as equivalent, overlap them.

Answer 1

Concepts and reason

The required concepts to solve these questions are effective resistance of the combinations of the resistor, power dissipated by the bulbs and potential across the bulbs.

Firstly, solve the circuit by using series and parallel combination of the resistor, calculate the current in the circuit and calculate power dissipated by the bulbs. Rank the bulbs based on the brightest to dimmest.

Fundamentals

The expression for equivalent resistance for series combination of resistors is,

${R_{\rm{s}}} = {R_1} + {R_2}$

Here, ${R_s}$ is the equivalent resistance of the bulb in series, ${R_1}$ is the resistance of the bulb and ${R_2}$ is the resistance of the other bulb.

The expression for parallel combination of resistors is,

${R_{\rm{p}}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$

Here, ${R_p}$ is the equivalent resistance of the bulb in parallel.

The expression for circuit in the circuit is.

$I = \frac{V}{R}$

Here, $R$ is the resistance and $V$ is the potential.

The expression for power dissipated by bulb is,

$P = {I^2}R$

Here, $I$ is the current in the circuit and $R$ is the resistance.

Diagram of bulbs is given as,

For bulb A and B.

Bulb A and B are connected in parallel. The expression for equivalent resistance for parallel connection is,

${R_{{\rm{AB}}}} = \frac{{{R_{\rm{A}}}{R_{\rm{B}}}}}{{{R_{\rm{A}}} + {R_{\rm{B}}}}}$

Substitute $R$ for ${R_{\rm{A}}}$ and ${R_{\rm{B}}}$ in the above equation.

$\begin{array}{c}\\{R_{{\rm{AB}}}} = \frac{{RR}}{{R + R}}\\\\ = \frac{R}{2}\\\end{array}$ …… (1)

Bulb D and E are connected in series. The expression for equivalent resistance of bulb D and E connected in series is,

${R_{{\rm{DE}}}} = {R_{\rm{D}}} + {R_{\rm{E}}}$

Substitute $R$ for ${R_{\rm{D}}}$ and ${R_{\rm{E}}}$ in the above equation.

$\begin{array}{c}\\{R_{{\rm{DE}}}} = R + R\\\\ = 2R\\\end{array}$ …… (2)

Bulb DE is parallel with bulb C. The expression for equivalent resistance for parallel connection is,

${R_{{\rm{CDE}}}} = \frac{{{R_{\rm{C}}}{R_{{\rm{DE}}}}}}{{{R_{\rm{C}}} + {R_{{\rm{DE}}}}}}$

Substitute $R$ for ${R_{\rm{C}}}$ and $2R$ for ${R_{{\rm{DE}}}}$ in the above equation.

$\begin{array}{c}\\{R_{{\rm{CDE}}}} = \frac{{{R_{\rm{C}}}{R_{{\rm{DE}}}}}}{{{R_{\rm{C}}} + {R_{{\rm{DE}}}}}}\\\\ = \frac{{R\left( {2R} \right)}}{{R + 2R}}\\\\ = \frac{{2{R^2}}}{{3R}}\\\\ = \frac{{2R}}{3}\\\end{array}$ …… (3)

The expression for current ${I_{{\rm{AB}}}}$ in the circuit is,

${I_{{\rm{AB}}}} = \frac{V}{{{R_{{\rm{AB}}}}}}$

Here, ${I_{{\rm{AB}}}}$ is the current through the bulb A and B, is the potential and ${R_{{\rm{AB}}}}$ is the resistance of bulb AB.

From the equation (1), ${R_{{\rm{AB}}}} = \frac{R}{2}$ .

Substitute $\frac{R}{2}$ for ${R_{{\rm{AB}}}}$ in the above equation.

${I_{{\rm{AB}}}} = \frac{{2V}}{R}$ …… (4)

A and B have equal resistance and they are connected in parallel so current divides equally.

Thus,

${I_{\rm{A}}} = \frac{V}{R}$ and

${I_{\rm{B}}} = \frac{V}{R}$ .

The expression for current ${I_{{\rm{DE}}}}$ in the circuit is,

${I_{{\rm{DE}}}} = \frac{V}{{{R_{{\rm{DE}}}}}}$

Here, ${I_{{\rm{DE}}}}$ is the current through the bulb D and E.

From the equation (2), ${R_{{\rm{DE}}}} = 2R$ .

Substitute $2R$ for ${R_{{\rm{DE}}}}$ in the above equation.

${I_{{\rm{DE}}}} = \frac{V}{{2R}}$ …… (5)

The expression for current ${I_{{\rm{CDE}}}}$ in the circuit is,

${I_{{\rm{CDE}}}} = \frac{V}{{{R_{{\rm{CDE}}}}}}$

Here, ${I_{{\rm{CDE}}}}$ is the current through the bulb C, D and E.

From the equation (3), ${R_{{\rm{CDE}}}} = \frac{{2R}}{3}$ .

Substitute $\frac{{2R}}{3}$ for ${R_{{\rm{CDE}}}}$ in the above equation.

${I_{{\rm{CDE}}}} = \frac{{3V}}{{2R}}$ …… (6)

The expression for power dissipated by bulb A is,

${P_{\rm{A}}} = {I_{\rm{A}}}^2{R_{\rm{A}}}$

Substitute $\frac{V}{R}$ for ${I_{\rm{A}}}$ and $R$ for ${R_{\rm{A}}}$ in above equation.

$\begin{array}{c}\\{P_{\rm{A}}} = {\left( {\frac{V}{R}} \right)^2}R\\\\ = {\frac{V}{R}^2}\\\end{array}$

Bulb A and B has same resistance and same current. They dissipate equal power. So

${P_{\rm{A}}} = {P_{\rm{B}}}$ . …… (7)

Substitute ${\frac{V}{R}^2}$ for ${P_{\rm{A}}}$ in the above equation.

${P_{\rm{B}}} = {\frac{V}{R}^2}$

The expression for power dissipated by bulb C is,

${P_{\rm{C}}} = {I_{\rm{C}}}^2{R_{\rm{C}}}$

There are three bulbs in that bulb C is in parallel with bulb DE.  $\frac{2}{3}$ of current pass through bulb C and $\frac{1}{3}$ of current pass through bulb DE.

From the equation (6),

Substitute $\frac{2}{3}{I_{{\rm{CDE}}}}$ for ${I_{\rm{C}}}$ and $R$ for ${R_{\rm{C}}}$ in the above equation.

${P_{\rm{C}}} = {\left( {\frac{2}{3}{I_{{\rm{CDE}}}}} \right)^2}R$

Substitute $\frac{{3V}}{{2R}}$ for ${I_{{\rm{CDE}}}}$ in above equation.

$\begin{array}{c}\\{P_{\rm{C}}} = {\left( {\frac{2}{3}\left( {\frac{{3V}}{{2R}}} \right)} \right)^2}R\\\\ = \frac{3}{2}\left( {\frac{{{V^2}}}{R}} \right)\\\end{array}$

Bulb D and E are connected in the series so power dissipated in the D and E are same.

${P_{\rm{D}}} = {P_{\rm{E}}}$ …… (8)

The expression for power dissipated by bulb D is,

${P_{\rm{D}}} = \frac{1}{3}{I_{{\rm{CDE}}}}^2{R_{\rm{D}}}$

Substitute $\frac{{3V}}{{2R}}$ for ${I_{{\rm{CDE}}}}$ and $R$ for ${R_{\rm{D}}}$ in the above equation.

$\begin{array}{c}\\{P_{\rm{D}}} = \frac{1}{3}{\left( {\frac{{3V}}{{2R}}} \right)^2}R\\\\ = \frac{3}{4}\frac{{{V^2}}}{R}\\\end{array}$

Substitute $\frac{3}{4}\frac{{{V^2}}}{R}$ for ${P_{\rm{D}}}$ in equation (8).

${P_{\rm{E}}} = \frac{3}{4}\frac{{{V^2}}}{R}$

Compare the power of bulbs. Power dissipated by bulb A, B, C, D, E are ${\frac{V}{R}^2}$ , ${\frac{V}{R}^2}$ , $\frac{3}{2}\left( {\frac{{{V^2}}}{R}} \right)$ , $\left( {\frac{3}{4}\frac{{{V^2}}}{R}} \right)$ , $\left( {\frac{3}{4}\frac{{{V^2}}}{R}} \right)$ .

Rank of the bulbs based on their brightness.

${\rm{C}} > {\rm{A}} = {\rm{B}} > {\rm{D}} = {\rm{E}}$

Bulb C has brightest than A and B which are equal and D and E are dimmest.

Ans:

Rank of the bulbs based on their brightness are ${\rm{C}} > {\rm{A}} = {\rm{B}} > {\rm{D}} = {\rm{E}}$ .