Chapter 7, Problem 34Q

To determine

The ranking from smallest to greatest of the given circuit, as shown in figure 1, for the configuration of resistances from A to F in terms of voltage across the bulb highlighted in red.

Answer to Problem 34Q

Ranking of the circuit from greatest to smallest, according to the voltage across the bulb shown in red in the list is as follows:

Rank 1: Circumstance F: Bulb-2: 20 Ω

Rank 2: Circumstance D: Bulb-2: 20 Ω

Rank 3: Circumstance B: Bulb-2: 10 Ω

Rank 3: Circumstance C: Bulb-3: 10 Ω

Rank 3: Circumstance E: Bulb-1: 10 Ω

Rank 4: Circumstance A: Bulb-1: 10 Ω

Explanation of Solution

Given information:

The given circuit is as shown in figure 1:

And the six circumstances for the arrangement of bulbs are as:

A: Bulb 1: 10 Ω; bulb2: 20 Ω; bulb 3: 30 Ω

B: Bulb 1: 10 Ω; bulb2: 10 Ω; bulb 3: 20 Ω

C: Bulb 1: 10 Ω; bulb2: 20 Ω; bulb 3: 10 Ω

D: Bulb 1: 10 Ω; bulb2: 20 Ω; bulb 3: 30 Ω

E.: Bulb 1: 10 Ω; bulb2: 10 Ω; bulb 3: 20 Ω

F: Bulb 1: 10 Ω; bulb2: 20 Ω; bulb 3: 10 Ω

For each situation, however, the battery and the wire connections in the circuit are always the same.

Concept Used:

1. Ohm’s law states that

$V=IR$

2. The resistance $\left(R\right)$ of a wire having resistivity, area and length as $\rho$, $A$ and $l$ respectively, is as:

   $R=\rho \frac{l}{A}$

   Where, $\rho$ is the resistivity of the conductor.

   l is the length of the conductor and

   A is the area of the conductor.

3. For the resistors connected in series, the net resistance is the algebraic sum of the individual resistance.

Calculation:

For the circuit shown in figure 1 let us consider that the voltage supplied by the battery as V.

Let R_A, R_B, R_C, R_D, R_E & R_F be the total resistance of the circuit under circumstances A, B, C, D, E & F respectively.

And let I_A, I_B, I_C, I_D, I_E & I_F be the currents flowing through the circuit under circumstances A, B, C, D, E & F respectively.

Therefore, the parameter values in each circumstance is as:

For circumstance A,

Total resistance under circumstance A

$\begin{array}{l}{R}_A=R_{Bulb1}+R_{Bulb2}+R_{Bulb3}\\ \Rightarrow R_A=10+20+30\because R_{Bulb1}=10\Omega ,R_{Bulb2}=20\Omega ,R_{Bulb3}=30\Omega \\ \Rightarrow R_A=60\Omega \end{array}$

The current flowing through the circuit under circumstances A, $I_A=\frac{V}{60}\text{A}$

Voltage across bulb 1: 10 Ω under circumstance A,

$\begin{array}{l}{V}_A=I_A{R}_{Bulb1}\\ \Rightarrow V_A=\frac{V}{60}\times 10\\ \Rightarrow V_A=\frac{V}{6}\text{Volts}\end{array}$

For circumstance B,

Total resistance under circumstance B,

$\begin{array}{l}{R}_B=R_{Bulb1}+R_{Bulb2}+R_{Bulb3}\\ \Rightarrow R_B=10+10+20\because R_{Bulb1}=10\Omega ,R_{Bulb2}=10\Omega ,R_{Bulb3}=20\Omega \\ \Rightarrow R_B=40\Omega \end{array}$

The current flowing through the circuit under circumstances B, $I_B=\frac{V}{40}\text{A}$

Voltage across bulb 2: 10 Ω under circumstance B,

$\begin{array}{l}{V}_B=I_B{R}_{Bulb2}\\ \Rightarrow V_B=\frac{V}{40}\times 10\\ \Rightarrow V_B=\frac{V}{4}\text{Volts}\end{array}$

For circumstance C,

Total resistance under circumstance C,

$\begin{array}{l}{R}_C=R_{Bulb1}+R_{Bulb2}+R_{Bulb3}\\ \Rightarrow R_C=10+20+10\because R_{Bulb1}=10\Omega ,R_{Bulb2}=20\Omega ,R_{Bulb3}=10\Omega \\ \Rightarrow R_C=40\Omega \end{array}$

The current flowing through the circuit under circumstances C, $I_C=\frac{V}{40}\text{A}$

Voltage across bulb 3: 10 Ω under circumstance C,

$\begin{array}{l}{V}_C=I_C{R}_{Bulb3}\\ \Rightarrow V_C=\frac{V}{40}\times 10\\ \Rightarrow V_C=\frac{V}{4}\text{Volts}\end{array}$

For circumstance D,

Total resistance under circumstance D,

$\begin{array}{l}{R}_D=R_{Bulb1}+R_{Bulb2}+R_{Bulb3}\\ \Rightarrow R_D=10+20+30\because R_{Bulb1}=10\Omega ,R_{Bulb2}=20\Omega ,R_{Bulb3}=30\Omega \\ \Rightarrow R_D=60\Omega \end{array}$

The current flowing through the circuit under circumstances D, $I_D=\frac{V}{60}\text{A}$

Voltage across bulb 2: 20 Ω under circumstance D,

$\begin{array}{l}{V}_D=I_D{R}_{Bulb2}\\ \Rightarrow V_D=\frac{V}{60}\times 20\\ \Rightarrow V_D=\frac{V}{3}\text{Volts}\end{array}$

For circumstance E,

Total resistance under circumstance E,

$\begin{array}{l}{R}_E=R_{Bulb1}+R_{Bulb2}+R_{Bulb3}\\ \Rightarrow R_E=10+10+20\because R_{Bulb1}=10\Omega ,R_{Bulb2}=10\Omega ,R_{Bulb3}=20\Omega \\ \Rightarrow R_E=40\Omega \end{array}$

The current flowing through the circuit under circumstances E, $I_E=\frac{V}{40}\text{A}$

Voltage across bulb 1: 10 Ω under circumstance E,

$\begin{array}{l}{V}_E=I_E{R}_{Bulb1}\\ \Rightarrow V_E=\frac{V}{40}\times 10\\ \Rightarrow V_E=\frac{V}{4}\text{Volts}\end{array}$

For circumstance F,

Total resistance under circumstance F,

$\begin{array}{l}{R}_F=R_{Bulb1}+R_{Bulb2}+R_{Bulb3}\\ \Rightarrow R_F=10+20+10\because R_{Bulb1}=10\Omega ,R_{Bulb2}=20\Omega ,R_{Bulb3}=10\Omega \\ \Rightarrow R_F=40\Omega \end{array}$

The current flowing through the circuit under circumstance F, $I_F=\frac{V}{40}\text{A}$

Voltage across bulb 2: 20 Ω under circumstance F,

$\begin{array}{l}{V}_F=I_F{R}_{Bulb1}\\ \Rightarrow V_F=\frac{V}{40}\times 20\\ \Rightarrow V_F=\frac{V}{2}\text{Volts}\end{array}$

Therefore, the ranking of voltages in the highlighted bulb for the configurations from A to F is:

$V_F<V_D<V_B=V_C=V_E<V_A$.

Conclusion:

Ranking of the circuit from greatest to smallest, according to the voltage across the bulb shown in red in the list is as follows:

Rank 1: Circumstance F: Bulb-2: 20 Ω

Rank 2: Circumstance D: Bulb-2: 20 Ω

Rank 3: Circumstance B: Bulb-2: 10 Ω

Rank 3: Circumstance C: Bulb-3: 10 Ω

Rank 3: Circumstance E: Bulb-1: 10 Ω

Rank 4: Circumstance A: Bulb-1: 10 Ω