In Fig. 27-5a, find the potential difference acrossR₂if$\mathit{\epsilon }\mathbf{=}\mathbf{12}\mathit{V}\mathbf{,}\mathbf{}{\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathit{\Omega }\mathbf{,}{\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathit{\Omega }\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{R}}_{\mathbf{3}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{\Omega }$

• a) The given emf of the battery in the circuit,$\epsilon =12V$.

The given resistances$R_1=3.0\Omega ,R_2=4.0\Omega ,R_3=5.0\Omega$

For the pair of resistors in a series connection, the current value remains the same for each resistor. Thus, the voltage drop across each resistor is divided equally as per the values of the resistances.

Formulae:

The voltage equation using Ohm’s law,

V=IR (i)

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

Kirchhoff’s voltage law,

(ii)

Now, as the value of current is the same throughout the closed loop, using equation (i) in equation (ii), we can get the current equation for the given capacitors as follows:

Thus, the potential difference across resistor 2 is given using the above value in equation (i) as follows:

Substitute the values in the above formula, and we get,

Hence, the value of the potential difference is 4.0V.