Question

 

Bulbs A, B, and C in the figure(Part A figure)are identical and the switch is an ideal conductor. How does closing the switch in the figure affect the potential difference? Check all that apply.

	The potential difference across A is unchanged.
	The potential difference across B drops to zero.
	The potential difference across A increases by 50%.
	The potential difference across B drops by 50%.

 

One more bulb is added to the circuit and the location of the switch is changed. The new circuit is shown in the figure.(Part B figure)Bulbs A, B, C, and D are identical and the switch is an ideal conductor. How does closing the switch in the figure affect the potential difference? Check all that apply.

	The potential difference across A increases.
	The potential difference across B doubles.
	The potential difference across B drops to zero.
	The potential difference across D is unchanged.

 

 

Answer 1

Concepts and reason

The problem deals with the concept of the ohms law, series resistance and parallel resistance as the effect of closing the switch is to be determined on the voltage.

Determine the net resistance of the circuit when the switch is open and determine the change in voltage across A, B, C and D. And then clos the switch and then determine the net resistance of the circuit and the change in voltage across A, B, C and D.

Fundamentals

According to ohm’s law the current flowing through a circuit is directly proportional to the voltage across the circuit and inversely proportional to the resistance.

The ohm’s law can be represented as:

$V = IR$

Here $I$ is the current, $R$ is the resistance and $V$ is the voltage.

When $n$ resistance ${R_1},{R_2},{R_3}.....{R_n}$ are in parallel in a circuit then the net resistance of the circuit will be:

$\frac{1}{{{R_p}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}} + ............. + \frac{1}{{{R_n}}}$

Here ${R_p}$ is the net resistance.

When $n$ resistance ${R_1},{R_2},{R_3}.....{R_n}$ are in series in a circuit then the net resistance of the circuit will be:

${R_s} = {R_1} + {R_2} + {R_3} + .......... + {R_n}$

Here ${R_s}$ is the net resistance.

Since the resistance in the bulbs A, B, C and D is equal. So to make the calculation easy suppose the value of each resistance is $50\Omega$ and suppose the voltage of the battery is $V = 83.33{\rm{V}}$ .

Now the resistance ${R_B}$ and ${R_C}$ are in series therefore the net resistance of these two resistance will be as:

${R_s} = {R_1} + {R_2} + {R_3} + .......... + {R_n}$

Substitute $50\Omega$ for ${R_B}$ and $50\Omega$ for ${R_C}$ as:

$\begin{array}{c}\\{R_{BC}} = {R_B} + {R_C}\\\\ = 50\Omega + 50\Omega \\\\ = 100\Omega \\\end{array}$

Now the resistance ${R_{BC}}$ and ${R_D}$ are in parallel therefore the net resistance will be as:

$\frac{1}{{{R_p}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}} + ............. + \frac{1}{{{R_n}}}$

Substitute $100\Omega$ for ${R_{BC}}$ and $50\Omega$ for ${R_D}$ as:

$\begin{array}{c}\\\frac{1}{{{R_{BCD}}}} = \frac{1}{{{R_{BC}}}} + \frac{1}{{{R_D}}}\\\\ = \frac{1}{{100\Omega }} + \frac{1}{{50\Omega }}\\\\ = \frac{{1 + 2}}{{100\Omega }}\\\\ = \frac{3}{{100\Omega }}\\\end{array}$

Rearrange the equation to determine ${R_{BCD}}$ as:

$\begin{array}{l}\\\frac{1}{{{R_{BCD}}}} = \frac{3}{{100\Omega }}\\\\{R_{BCD}} = \frac{{100\Omega }}{3}\\\\{R_{BCD}} = 33.33\Omega \\\end{array}$

Now the resistance ${R_{BCD}}$ and ${R_A}$ are in series therefore the net resistance of these two resistance will be as:

${R_s} = {R_1} + {R_2} + {R_3} + .......... + {R_n}$

Substitute $33.33\Omega$ for ${R_{BCD}}$ and $50\Omega$ for ${R_A}$ as:

$\begin{array}{c}\\{R_{ABCD}} = {R_{BCD}} + {R_A}\\\\ = 33.33\Omega + 50\Omega \\\\ = 83.33\Omega \\\end{array}$

Therefore the net resistance of the circuit when the switch is open is $83.33\Omega$ .

From the ohm’s law determine the current from the circuit as:

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

Substitute the value of voltage $V = 83.33{\rm{V}}$ and the net resistance of the circuit $R = 83.33\Omega$ as:

$\begin{array}{c}\\I = \frac{V}{R}\\\\ = \frac{{83.33{\rm{V}}}}{{83.33\Omega }}\\\\ = 1{\rm{A}}\\\end{array}$

Therefore again from the ohm’s law the voltage across the resistance ${R_A}$ is as:

$V = IR$

Substitute the value of current $I = 1{\rm{A}}$ and the value of resistance A ${R_A} = 50\Omega$ as:

$\begin{array}{c}\\{V_A} = I{R_A}\\\\ = \left( {1{\rm{A}}} \right)\left( {50\Omega } \right)\\\\ = 50{\rm{V}}\\\end{array}$

Since in this case the switch is closed so no current will flow from bulb C as the current will by pass from the alternative switch path. So there is no use of presence of resistance ${R_C}$ in the circuit in this scenario.

Now the resistance ${R_B}$ and ${R_D}$ are in parallel therefore the net resistance will be as:

$\frac{1}{{{R_p}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}} + ............. + \frac{1}{{{R_n}}}$

Substitute $50\Omega$ for ${R_B}$ and $50\Omega$ for ${R_D}$ as:

$\begin{array}{c}\\\frac{1}{{{R_{BD}}}} = \frac{1}{{{R_B}}} + \frac{1}{{{R_D}}}\\\\ = \frac{1}{{50\Omega }} + \frac{1}{{50\Omega }}\\\\ = \frac{{1 + 1}}{{50\Omega }}\\\\ = \frac{2}{{50\Omega }}\\\end{array}$

Rearrange the equation to determine ${R_{BD}}$ as:

$\begin{array}{l}\\\frac{1}{{{R_{BD}}}} = \frac{2}{{50\Omega }}\\\\{R_{BD}} = \frac{{50\Omega }}{2}\\\\{R_{BD}} = 25\Omega \\\end{array}$

Now the resistance ${R_{BD}}$ and ${R_A}$ are in series therefore the net resistance of these two resistance will be as:

${R_s} = {R_1} + {R_2} + {R_3} + .......... + {R_n}$

Substitute $25\Omega$ for ${R_{BD}}$ and $50\Omega$ for ${R_A}$ as:

$\begin{array}{c}\\{R_{ABD}} = {R_{BD}} + {R_A}\\\\ = 25\Omega + 50\Omega \\\\ = 75\Omega \\\end{array}$

Therefore the net resistance of the circuit when the switch is closed is $75\Omega$ .

From the ohm’s law determine the current from the circuit as:

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

Substitute the value of voltage $V = 83.33{\rm{V}}$ and the net resistance of the circuit $R = 75\Omega$ as:

$\begin{array}{c}\\I = \frac{V}{R}\\\\ = \frac{{83.33{\rm{V}}}}{{75\Omega }}\\\\ = 1.11{\rm{A}}\\\end{array}$

Therefore again from the ohm’s law the voltage across the resistance ${R_A}$ is as:

$V = IR$

Substitute the value of current $I = 1.11{\rm{A}}$ and the value of resistance A ${R_A} = 50\Omega$ as:

$\begin{array}{c}\\{V_A} = I{R_A}\\\\ = \left( {1.11{\rm{A}}} \right)\left( {50\Omega } \right)\\\\ = 55.50{\rm{V}}\\\end{array}$

Therefore the voltage in the bulb A has increased from $50{\rm{V}}$ to $55.50{\rm{V}}$ when the switch is closed.

Ans:

The potential difference across A increases as the voltage in the bulb A has increased from $50{\rm{V}}$ to $55.50{\rm{V}}$ when the switch is closed.