Question

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?

Short answer

The potential at point Q is${\mathrm{V}}_{\mathrm{Q}}=-10\mathrm{V}$.

Step-by-step solution

Given

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$\mathrm{P}=100\mathrm{V}$

Determining the concept

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.

Determining the potential at point Q

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