91Ó°ÊÓ

Consider the given value of the voltage and the resistances:

$\mathrm{Î\mu }=6.0V$

${\mathrm{R}}_1=6.0\mathrm{Î\copyright }$
${\mathrm{R}}_2=12.0\mathrm{Î\copyright }$

${\mathrm{R}}_3=4.0\mathrm{Î\copyright }$

$R_4=3.00\mathrm{Î\copyright }$

$R_5=5.00\mathrm{Î\copyright }$

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:

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

Consider the voltage across R₄₅=12.0V

Hence, current through R₄and R₅ is given by:

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

Solve for the voltage across R₅ as follows:

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

The potential difference across R₅is V_R5=7.5V.