Question

What is the value of resistor R inthe figure below, in which I = 8 A and ΔV = 34 V?

What is the value of resistor R inthe figure below, in which I = 8 A and triangle V = 34 V?

Answer 1

Concepts and reason

The main concept used to solve the problem is ohm’s law.

Initially, calculate the equivalent resistance using the parallel and series combination. Later, use the ohm’s law. Finally, calculate the magnitude of resistance by substituting the respective values.

Fundamentals

The ohm’s law is expressed as follows:

$V = IR$

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

The equivalent resistance in a parallel combination is calculated as follows:

$\frac{1}{{{R_{{\rm{eq}}}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}....\frac{1}{{{R_n}}}$

Here, ${R_{{\rm{eq}}}}$ is the equivalent resistance, ${R_1}$ is the resistance1, ${R_2}$ is the resistance 2, ${R_3}$ is the resistance 3, and ${R_n}$ is the resistance3.

According to Ohm’s law, equivalent resistance is,

${R_{{\rm{eq}}}} = \frac{V}{I}$

Substitute 34 V for V and 8 A for I in expression ${R_{{\rm{eq}}}} = \frac{V}{I}$.

$\begin{array}{c}\\{R_{{\rm{eq}}}} = \frac{{34\,{\rm{V}}}}{{8\,{\rm{A}}}}\\\\ = 4.25\,\Omega \\\end{array}$

The equivalent resistance in a parallel combination is calculated as follows:

$\frac{1}{{{R_{{\rm{eq}}}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}....\frac{1}{{{R_n}}}$

Here, ${R_{{\rm{eq}}}}$ is the equivalent resistance, ${R_1}$ is the resistance1, ${R_2}$ is the resistance 2, ${R_3}$ is the resistance 3, and ${R_n}$ is the resistance 3.

For the circuit, the equivalent resistance is calculated by using the expression,

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

Substitute $4.25\,\Omega$ for ${R_{{\rm{eq}}}}$, $10\,\Omega$ for ${R_1}$, $15\,\Omega$ for ${R_2}$, and $R$ for ${R_3}$ in expression $\frac{1}{{{R_{{\rm{eq}}}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}$.

$\begin{array}{c}\\\frac{1}{{4.25\,\Omega }} = \frac{1}{{10\,\Omega }} + \frac{1}{{15\,\Omega }} + \frac{1}{R}\\\\\frac{1}{R} = \left( {\frac{1}{{4.25}}} \right) - \left( {\frac{1}{6}} \right)\\\\R = 14.57\,\Omega \\\end{array}$

Ans:

The value of resistor R is $14.6\,\Omega$.