A circuit is as shown in the figure. The current from A to B through…

The correct answer is Option (2) → +250 mA

By Kirchhoff’s junction rule, the sum of the currents entering junction A equals the sum of the currents leaving it. Taking B as zero potential, each branch current is the potential difference across its resistor divided by the resistance.

From the figure, the bottom wire is at the potential of B:

• The 10 V cell has its positive terminal at the top, so the left end of the \( 10\,\Omega \) resistor is at +10 V.

• The 5 V cell has its negative terminal at the top, so the right end of the \( 15\,\Omega \) resistor is at −5 V.

Let the potential of A be \( V \) (potentials in volts, resistances in ohms). The currents flowing into A from the two sides add up to the current from A to B:

    \( \displaystyle \frac{10-V}{10}+\frac{-5-V}{15}=\frac{V}{10} \)

    \( 3\,(10-V)+2\,(-5-V)=3V \)

    \( 20-5V=3V \)

    \( \Rightarrow\; V=2.5\,\mathrm{V} \)

The current from A to B through the \( 10\,\Omega \) resistor is:

    \( \begin{aligned} I_{AB} &= \frac{V_A-V_B}{10\,\Omega} \\ &= \frac{2.5\,\mathrm{V}}{10\,\Omega} \\ &= 0.25\,\mathrm{A} \\ &= 250\,\mathrm{mA} \end{aligned} \)

The positive sign shows that the current flows from A to B.