Question

a) For the circuit shown in the figure find the current through each resistor.

b) For the circuit shown in the figure find the potential difference across each resistor.

Answer 1

Concepts and reason

The concept required to solve the given problem is series and parallel combination of resistors and ohm’s law.

First use the formula for parallel and series combination of resistors to solve the equivalent resistance of the circuit.

Then, use Ohm’s law to calculate the current and potential drop across the resistors.

Fundamentals

The ohm’s law states that the current flowing through the conductor is directly proportional to the potential difference across its ends. Thus, the numerical form of the law is,

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

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

The resistance of the resistors in series is calculated by,

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

Here, ${R_1}$ , ${R_2}$ , ${R_3}$ ,… ${R_n}$ are the resistance of the resistors.

The resistance of the resistors in parallel is calculated by,

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

Here, ${R_1}$ , ${R_2}$ , ${R_3}$ ,… ${R_n}$ are the resistance of the resistors.

Across series combination of resistors current across each resistor is same and is equal to the current flowing through the circuit.

Across parallel combination of resistors voltage across each resistor is same and is equal to the voltage of the battery.

(a)

The following figure shows a circuit diagram with resistances ${R_1} = 4.00{\rm{ }}\Omega ,{R_2} = 6.00{\rm{ }}\Omega ,{R_3} = 8.00{\rm{ }}\Omega {\rm{ and }}{R_4} = {R_5} = 24.0{\rm{ }}\Omega$ and a battery with voltage $24{\rm{ V}}$ .

Since the resistors, ${R_3}{\rm{ and }}{R_4}$ are connected in parallel, the equivalent resistance of the resistors, ${R_3}{\rm{ and }}{R_4}$ is,

$\frac{1}{{{R_{34}}}} = \frac{1}{{{R_3}}} + \frac{1}{{{R_4}}}$

Substitute $8{\rm{ }}\Omega$ for ${R_3}$ and $24{\rm{ }}\Omega$ for ${R_4}$ in the above equation.

$\begin{array}{c}\\\frac{1}{{{R_{34}}}} = \frac{1}{{8{\rm{ }}\Omega }} + \frac{1}{{24{\rm{ }}\Omega }}\\\\{R_{34}} = 6{\rm{ }}\Omega \\\end{array}$

Since the resistors, ${R_{34}}{\rm{ and }}{R_2}$ are in series, the equivalent resistance of resistors, ${R_{34}}{\rm{ and }}{R_2}$ is,

${R_{234}} = {R_2} + {R_{34}}$

Substitute ${\rm{6 }}\Omega$ for ${R_2}$ and $6{\rm{ }}\Omega$ for ${R_{34}}$ in the above equation.

$\begin{array}{c}\\{R_{234}} = 6{\rm{ }}\Omega + 6{\rm{ }}\Omega \\\\ = 12{\rm{ }}\Omega \\\end{array}$

Since the resistors, ${R_{234}}{\rm{ and }}{R_5}$ are connected in parallel, the equivalent resistance of the resistors, ${R_{234}}{\rm{ and }}{R_5}$ is,

$\frac{1}{{{R_{2345}}}} = \frac{1}{{{R_{234}}}} + \frac{1}{{{R_5}}}$

Substitute $12{\rm{ }}\Omega$ for ${R_{234}}$ and $24{\rm{ }}\Omega$ for ${R_5}$ in the above equation.

$\begin{array}{c}\\\frac{1}{{{R_{2345}}}} = \frac{1}{{12{\rm{ }}\Omega }} + \frac{1}{{24{\rm{ }}\Omega }}\\\\{R_{2345}} = 8{\rm{ }}\Omega \\\end{array}$

Since the resistors ${R_1}{\rm{ and }}{R_{2345}}$ are in series with each other, the equivalent resistance of the circuit is,

${R_{eq}} = {R_1} + {R_{2345}}$

Substitute $4{\rm{ }}\Omega$ for ${R_1}$ and $8{\rm{ }}\Omega$ for ${R_{2345}}$ in the above equation.

$\begin{array}{c}\\{R_{eq}} = \left( {4{\rm{ }}\Omega } \right) + \left( {8{\rm{ }}\Omega } \right)\\\\ = 12{\rm{ }}\Omega \\\end{array}$

The current flowing through the circuit will be given by ohm’s law.

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

Substitute $24{\rm{ V}}$ for $V$ and $12{\rm{ }}\Omega$ for ${R_{2345}}$ in the above equation.

$\begin{array}{c}\\I = \frac{{24{\rm{ V}}}}{{{\rm{12 }}\Omega }}\\\\ = 2{\rm{ A}}\\\end{array}$

The current through the resistor ${R_1}$ will be equal to the current running in the circuit.

${I_1} = 2{\rm{ A}}$

The current through the equivalent resistance ${R_{2345}}$ will also be equal to $2{\rm{ A}}$ .

Hence, ${I_{2345}} = 2{\rm{ A}}$

The voltage through the resistor ${R_1}$ is,

${V_1} = {I_1}{R_1}$

Substitute $2{\rm{ A}}$ for ${I_1}$ and $4{\rm{ }}\Omega$ for ${R_1}$ in the above equation.

$\begin{array}{c}\\{V_1} = \left( {2{\rm{ A}}} \right)\left( {4{\rm{ }}\Omega } \right)\\\\ = 8{\rm{ V}}\\\end{array}$

The potential difference across ${R_{2345}}$ will be,

${V_{2345}} = \left( {24{\rm{ V}}} \right) - {V_1}$

Substitute $8{\rm{ V}}$ for ${V_1}$ in the above equation.

$\begin{array}{c}\\{V_{2345}} = \left( {24{\rm{ V}}} \right) - \left( {8{\rm{ V}}} \right)\\\\ = 16{\rm{ V}}\\\end{array}$

Since the resistors ${R_5}{\rm{ and }}{R_{234}}$ are in parallel to each other, the voltage drop across ${R_5}$ and across the equivalent resistance ${R_{234}}$ will be the same as ${V_{2345}}$ .

The voltage drop across the resistor, ${R_5}$ is,

${V_5} = 16{\rm{ V}}$

The current across the resistor, ${R_5}$ is,

${I_5} = \frac{{{V_5}}}{{{R_5}}}$

Substitute $16{\rm{ V}}$ for ${V_5}$ and $24{\rm{ }}\Omega$ for ${R_5}$ in the above equation.

$\begin{array}{c}\\{I_5} = \frac{{16{\rm{ V}}}}{{24{\rm{ }}\Omega }}\\\\ = 0.667{\rm{ A}}\\\end{array}$

The voltage drop across the resistor, ${R_{234}}$ is,

${V_{234}} = 16{\rm{ V}}$

The current across the equivalent resistor, ${R_{234}}$ is,

${I_{234}} = \frac{{{V_{234}}}}{{{R_{234}}}}$

Substitute $16{\rm{ V}}$ for ${V_{234}}$ and $12{\rm{ }}\Omega$ for ${R_{234}}$ in the above equation.

$\begin{array}{c}\\{I_{234}} = \frac{{16{\rm{ V}}}}{{12{\rm{ }}\Omega }}\\\\ = 1.33{\rm{ A}}\\\end{array}$

The current ${I_2}$ through the resistor, ${R_{234}}$ will be equal to ${I_{234}}$ .

Thus, ${I_2} = 1.33{\rm{ A}}$

The current ${I_{34}}$ through the equivalent resistor, ${R_{34}}$ will be equal to ${I_2}$ .

Thus, ${I_{34}} = 1.33{\rm{ A}}$

The voltage drop across ${R_{34}}$ will be,

${V_{34}} = {I_{34}}{R_{34}}$

Substitute $1.33{\rm{ A}}$ for ${I_{34}}$ and $6{\rm{ }}\Omega$ for ${R_{34}}$ in the above equation.

$\begin{array}{c}\\{V_{34}} = \left( {1.33{\rm{ A}}} \right)\left( {6{\rm{ }}\Omega } \right)\\\\ = 7.98{\rm{ V}}\\\end{array}$

The potential drop across ${R_3}{\rm{ and }}{R_4}$ will be equal to ${V_{34}}$ since the two resistors ${R_3}{\rm{ and }}{R_4}$ are parallel. Thus, the voltage drop across ${R_3}$ is,

$\begin{array}{c}\\{V_3} = {V_{34}}\\\\ = 7.98{\rm{ V}}\\\end{array}$

The current across the resistor ${R_3}$ will be given by,

${I_3} = \frac{{{V_3}}}{{{R_3}}}$

Substitute $7.98{\rm{ V}}$ for ${V_3}$ and $8{\rm{ }}\Omega$ for ${R_3}$ in the above equation.

$\begin{array}{c}\\{I_3} = \frac{{7.98{\rm{ V}}}}{{8{\rm{ }}\Omega }}\\\\ = 0.998{\rm{ A}}\\\end{array}$

The voltage drop across ${R_4}$ is,

$\begin{array}{c}\\{V_4} = {V_{34}}\\\\ = 7.98{\rm{ V}}\\\end{array}$

The current across the resistor ${R_4}$ will be given by,

${I_4} = \frac{{{V_4}}}{{{R_4}}}$

Substitute $7.98{\rm{ V}}$ for ${V_3}$ and $24{\rm{ }}\Omega$ for ${R_4}$ in the equation ${I_4} = \frac{{{V_4}}}{{{R_4}}}$ .

$\begin{array}{c}\\{I_4} = \frac{{7.98{\rm{ V}}}}{{{\rm{24 }}\Omega }}\\\\ = 0.333{\rm{ A}}\\\end{array}$

The voltage through the resistor ${R_1}$ is,

${V_1} = {I_1}{R_1}$

Substitute $2{\rm{ A}}$ for ${I_1}$ and $4{\rm{ }}\Omega$ for ${R_1}$ in the above equation.

$\begin{array}{c}\\{V_1} = \left( {2{\rm{ A}}} \right)\left( {4{\rm{ }}\Omega } \right)\\\\ = 8{\rm{ V}}\\\end{array}$

Since the resistors ${R_5}{\rm{ and }}{R_{234}}$ are in parallel to each other, the voltage drop across ${R_5}$ and across the equivalent resistance ${R_{234}}$ will be the same as ${V_{2345}}$ .

The voltage drop across the resistor, ${R_5}$ is,

${V_5} = 16{\rm{ V}}$

The potential drop across ${R_3}{\rm{ and }}{R_4}$ will be equal to ${V_{34}}$ since the tow resistors ${R_3}{\rm{ and }}{R_4}$ are parallel.

The voltage drop across ${R_3}$ ,

$\begin{array}{c}\\{V_3} = {V_{34}}\\\\ = 7.98{\rm{ V}}\\\end{array}$

The voltage drop across ${R_4}$ ,

$\begin{array}{c}\\{V_4} = {V_{34}}\\\\ = 7.98{\rm{ V}}\\\end{array}$

The voltage across resistors ${R_2}$ is,

${V_2} = {I_2}{R_2}$

Substitute $1.33{\rm{ A}}$ for ${I_2}$ and $6{\rm{ }}\Omega$ for ${R_1}$ in the above equation.

$\begin{array}{c}\\{V_2} = \left( {{\rm{1}}{\rm{.33 A}}} \right)\left( {{\rm{6 }}\Omega } \right)\\\\ = 7.98{\rm{ V}}\\\end{array}$

Ans: Part a

The current through the resistors ${R_1},{R_2},{R_3},{R_4}{\rm{ and }}{R_5}$ are 2 A, 1.33 A, 0.998 A, 0.333 A, and 0.667 A respectively.

Part b

The potential differences across the resistors ${R_1},{R_2},{R_3},{R_4}{\rm{ and }}{R_5}$ are 8.00 V, 7.98 V, 7.98 V, 7.98 V, and 16 V respectively.