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?

Consider the given value of the voltage and the resistances:

$\epsilon =6.0V$

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

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

$R_4=3.00\Omega$

$R_5=5.00\Omega$

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

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.