Question

For the circuit shown in the figure both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.55 A.

1.What does the voltmeter read?

2.What is the emf ε of the battery?

Answer 1

Concepts and reason

The concepts required to solve the given problem are series and parallel combination of resistors.

First, calculate the equivalent resistance of the resistors connected in series and parallel combination. Later the total magnitude of current through the circuit can be calculated by using the Ohm’s law and finally the reading across the voltmeter and emf of the battery will be calculate.

Fundamentals

Ohm’s law- It states that electric current flowing through a metallic wire is directly proportional to the potential difference V, across its ends provided its temperature remains the same

Resistors in series: When several resistors are joined in series, the resistance of the combination is equal to the sum of their individual resistances.

${R_{{\rm{eqv}}}} = {R_1} + {R_2} + {R_3}.......{R_{\rm{n}}}$

Here, ${R_{{\rm{eqv}}}}$ is the equivalent resistance, ${R_1}$ , R₂, and R_n are n^th number of resistors connected in series combination.

Resistors in parallel: When several resistors are joined in parallel, the reciprocal of the equivalent resistance of a group of resistance joined in parallel is equal to the sum of the reciprocal of the individual resistances.

$\frac{1}{{{R_{{\rm{eqv}}}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}.......\frac{1}{{{R_{\rm{n}}}}}$

Here, ${R_{{\rm{eqv}}}}$ is the equivalent resistance, ${R_1}$ , R₂, and R_n are n^th number of resistors connected in parallel combination.

(1)

Calculate the equivalent resistance of the three parallel resistors.

The equivalent resistance of the resistors connected in parallel combination is,

$\frac{1}{{{R_{{\rm{eqv}}}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}$

Here, ${R_{{\rm{eqv}}}}$ is the equivalent resistance, ${R_1}$ , R₂, and R₃ are resistors connected in parallel combination.

Substitute $25.0{\rm{ }}\Omega$ for ${R_1}$ , $15.0{\rm{ }}\Omega$ for ${R_2}$ , and $\left( {15.0{\rm{ }}\Omega + 10.0{\rm{ }}\Omega } \right)$ for ${R_3}$ , and solve for ${R_{{\rm{eqv}}}}$ .

$\begin{array}{c}\\\frac{1}{{{R_{{\rm{eqv}}}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}\\\\ = \frac{1}{{25.0{\rm{ }}\Omega }} + \frac{1}{{15.0{\rm{ }}\Omega }} + \frac{1}{{15.0{\rm{ }}\Omega {\rm{ + 10}}{\rm{.0 }}\Omega }}\\\\ = \frac{{11.0{\rm{ }}\Omega }}{{75.0{\rm{ }}\Omega }}\\\\{R_{{\rm{eqv}}}} = 6.82{\rm{ }}\Omega \\\end{array}$

Calculate the equivalent resistance of resistors connected in series combination.

Resistors R₄, R₅ and ${R_{{\rm{eqv}}}}$ are connected in series combination. So, their equivalent resistance is,

$R = {R_4} + {R_5} + {R_{{\rm{eqv}}}}$

Here, R is the final equivalent resistance of the circuit, and R₄, R₅ and ${R_{{\rm{eqv}}}}$ are resistances of the resistors connected in series.

Substitute $45.0{\rm{ }}\Omega$ for ${R_4}$ , $35.0{\rm{ }}\Omega$ for ${R_5}$ and $6.82{\rm{ }}\Omega$ for ${R_{{\rm{eqv}}}}$ .

$\begin{array}{c}\\R = {R_4} + {R_5} + {R_{{\rm{eqv}}}}\\\\ = \left( {45.0{\rm{ }}\Omega } \right) + \left( {35.0{\rm{ }}\Omega } \right) + \left( {6.82{\rm{ }}\Omega } \right)\\\\ = 86.82{\rm{ }}\Omega \\\end{array}$

Calculate the voltage drop across the $25.0{\rm{ }}\Omega$ resistor.

The voltage drop across the $25.0{\rm{ }}\Omega$ resistor is,

${V_{25}} = {I_1}{R_1}$

Here, I₁ is the current which ammeter reads, and R₁ is the resistance.

Substitute 1.55 A for I₁ and $25.0{\rm{ }}\Omega$ for R₁ as follows:

$\begin{array}{c}\\{V_{25}} = {I_1}{R_1}\\\\ = \left( {1.55{\rm{ A}}} \right)\left( {25.0{\rm{ }}\Omega } \right)\\\\ = 38.75{\rm{ V}}\\\end{array}$

Since resistors ${R_{\rm{2}}}$ and ${R_{\rm{3}}}$ are connected in parallel, so the potential drop across these resistors will remain same.

The current through resistor $\left( {15.0{\rm{ }}\Omega } \right)$ is,

${I_{15}} = \left( {\frac{{{V_{25}}}}{{{R_2}}}} \right)$

Here, ${V_{25}}$ is the voltage drop across $25.0{\rm{ }}\Omega$ and R₂ is the resistor.

Substitute 38.75 V for ${V_{25}}$ and $15.0{\rm{ }}\Omega$ for R₂.

$\begin{array}{c}\\{I_{15}} = \left( {\frac{{{V_{25}}}}{{{R_2}}}} \right)\\\\ = \left( {\frac{{38.75{\rm{ V}}}}{{15.0{\rm{ }}\Omega }}} \right)\\\\ = 2.58{\rm{ A}}\\\end{array}$

The current through the $\left( {15.0{\rm{ }}\Omega + 10.0{\rm{ }}\Omega } \right)$ is,

${I_{25}} = \frac{{{V_{25}}}}{{25.0{\rm{ }}\Omega }}$

Substitute 38.75 V for ${V_{25}}$ ,

$\begin{array}{c}\\{I_{25}} = \frac{{38.75{\rm{ V}}}}{{25.0{\rm{ }}\Omega }}\\\\ = 1.55{\rm{ A}}\\\end{array}$

Therefore, the total current in the circuit is,

$I = {I_1} + {I_{15}} + {I_{25}}$

Substitute 1.55 A for I₁, 2.58 A for ${I_{15}}$ , and 1.55 A for ${I_{25}}$ ,

$\begin{array}{c}\\I = {I_1} + {I_{15}} + {I_{25}}\\\\ = 1.55{\rm{ A + 2}}{\rm{.58 A + 1}}{\rm{.55 A}}\\\\{\rm{ = 5}}{\rm{.68 A}}\\\end{array}$

The potential difference across the $\left( {45.0{\rm{ }}\Omega } \right)$ is,

${V_{45}} = I{R_4}$

Substitute 5.68 A for I and $45.0{\rm{ }}\Omega$ for R₄,

$\begin{array}{c}\\{V_{45}} = \left( {5.68{\rm{ A}}} \right)\left( {45.0{\rm{ }}\Omega } \right)\\\\ = 255.6{\rm{ V}}\\\end{array}$

The emf of the battery is,

$\varepsilon {\rm{ = }}RI$

Here, $\varepsilon$ is the emf of the battery, R is the equivalent resistance of the whole circuit, and I is the magnitude of total current.

Substitute $86.82{\rm{ }}\Omega$ for R and $5.68{\rm{ A}}$ for I.

$\begin{array}{c}\\\varepsilon {\rm{ = }}RI\\\\ = \left( {86.82{\rm{ }}\Omega } \right)\left( {5.68{\rm{ A}}} \right)\\\\ = 493.138{\rm{ V}}\\\end{array}$

Ans: Part 1

The voltmeter reads 256 V.

Part 2

The emf of the battery is 493 V.