Question

In the circuit shown in the figure below, the magnitude and direction of the current will be

• A. $2.3Amp$ from A to B via E
• B. $2.3Amp$ from B to A via E
• C. $1.0Amp$ from B to A via E
• D. $1.0Amp$ from A to B via E

Answer 1

The correct option is A $2.3Amp$ from A to B via E

The voltage of the battery $V_1=10V$ and that of the battery $V_2=4V$ get added as they are connected in series with same polarity.

Net emf $V_{net}=10V+4V=14V$

The resistances $1\Omega ,3\Omega$ and $2\Omega$ are connected in series.

Total resistance $R_{series}=1+3+2=6\Omega$

By Ohm's law,

$V_{net}=I{R}_{series}$

$\Rightarrow I=\frac{V_{net}}{R_{series}}$

Hence, Current $I=\frac{14}{6}=2.3A$

For the direction of current, we can see net emf is in the same sense of the individual batteries because $10V$ battery and $4V$ battery are connected back-to-back. So $2.3A$ current will flow in the direction $A$ to $B$ via $E$.