Question

In Fig. 27-82, an ideal battery of emf $\mathbf{\epsilon }\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{V}$is connected to a network of resistances${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{\Omega }$, ${\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{\Omega }$,${\mathbf{R}}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\Omega }$,${\mathbf{R}}_{\mathbf{4}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\Omega }$and${\mathbf{R}}_{\mathbf{5}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\Omega }$. What is the potential difference across resistance 5?

Short answer

The potential difference across R₅is V_R5=7.5 V

Step-by-step solution

Determine the given quantities

Consider the given value of the voltage and the resistances:

$\epsilon =6.0V$

${\mathrm{R}}_1=6.0\Omega$
${\mathrm{R}}_2=12.0\Omega$

${\mathrm{R}}_3=4.0\Omega$

$R_4=3.00\Omega$

$R_5=5.00\Omega$

Determine the concept of Ohm’s Law

Here, we need to use the equation of Ohm’s law and the equivalent resistance of series combination. As the same current flows through the resistors connected in a series, and find the potential difference across an individual resistor.

For series connection the equivalent resistance is as follows:

R_eq= R_A+R_B

By Ohm’s law consider the formula as:

V=IR

Determine the potential different across the resistance as:

Equivalent of R₄ and R₅ is calculated as:

$\begin{array}{l}{R}_5=R_4+R_5\\ 3.00+5.00\\ =8.00\Omega \end{array}$

Consider the voltage across R₄₅=12.0V

Hence, current through R₄and R₅ is given by:

$\begin{array}{l}i=\frac{\epsilon }{R_{45}}\\ =\frac{12.0}{8.00}\\ =1.50A\\ \end{array}$

Solve for the voltage across R₅ as follows:

$\begin{array}{l}{V}_{R5}=i\times R_5\\ =1.50\times 5.00\\ =7.50V\end{array}$

The potential difference across R₅is V_R5=7.5V.