In Figure, the ideal batteries have emfs ${\mathbf{\epsilon }}_{\mathbf{1}}$=150 V and${\mathbf{\epsilon }}_{\mathbf{2}}$=50 V and the resistances are${\mathbf{R}}_{\mathbf{1}}$=3.0 Ω and${\mathbf{R}}_{\mathbf{2}}$=2.0 Ω. If the potential at Pis 100 V, what is it at Q?

Emf${\mathrm{\epsilon }}_1=150\mathrm{V}$

Emf${\mathrm{\epsilon }}_2=50\mathrm{V}$

Resistance${\mathrm{R}}_1=3\mathrm{\Omega }$

Resistance${\mathrm{R}}_2=2\mathrm{\Omega }$

Potential at point$\mathrm{P}=100\mathrm{V}$

Use the Kirchhoff’s loop law to find the current in the circuit. Then, from the equation obtained from Kirchhoff’s loop law and the current, write the relation between potential at P and Q. Then, inserting the values, get potential at point Q

Kirchhoff's loop rule states that the sum of all the electric potential differences around a loop is zero.

Formulae are as follow:

$\mathrm{V}=\mathrm{IR}$

Where, I is current, V is voltage, R is resistance.

Applying Kirchhoff’s loop law to the given circuit,

$$\begin{gathered} {\mathrm{\epsilon }}_1-{\mathrm{IR}}_1-{\mathrm{\epsilon }}_2-{\mathrm{IR}}_2=0 \\ 150-\mathrm{I}\left({\mathrm{R}}_1+{\mathrm{R}}_2\right)-50=0 \\ 100-5\mathrm{I}=0 \\ \mathrm{I}=20\mathrm{A} \end{gathered}$$

The potential at point Q is given by,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{P}}={\mathrm{V}}_{\mathrm{Q}}+150\mathrm{V}-\left(2\mathrm{\Omega }\right)\mathrm{I} \\ {\mathrm{V}}_{\mathrm{Q}}={\mathrm{V}}_{\mathrm{P}}-150\mathrm{V}+\left(2\mathrm{\Omega }\right)\mathrm{I} \\ {\mathrm{V}}_{\mathrm{Q}}=100\mathrm{V}-150\mathrm{V}+\left(2\mathrm{\Omega }\right)\left(20\mathrm{A}\right) \\ {\mathrm{V}}_{\mathrm{Q}}=-10\mathrm{V} \end{gathered}$$

Hence, the potential at point Q is${\mathrm{V}}_{\mathrm{Q}}=-10\mathrm{V}$

Therefore, by using the Kirchhoff’s loop law get the potential at point Q