Question

In the circuit shown in figure, the magnitude and the direction of the flow of current respectively would be

• A. $\frac{7}{2}\text{A}$ from $b$ to $a$ via $e$
• B. $1\text{A}$ from $a$ to $b$ via $e$
• C. $1\text{A}$ from $b$ to $a$ via $e$
• D. $\frac{7}{2}\text{A}$ from $a$ to $b$ via $e$

Answer 1

The correct option is B $1\text{A}$ from $a$ to $b$ via $e$

The circuit forms a closed loop therefore the algebraic sum of all the potential differences along the closed loop will be zero according to KVL,


$\sum \Delta V=0$

The potential drop across the resistors will be ${}^{\prime }i{R}^{\prime }$ in the direction of current flow.

$\Rightarrow -i\left(1\right)+10-4-2i-3i=0$

$\therefore i=1$

The direction of current will be dictated by the higher emf battery, because it will dominate the circuit.

$\Rightarrow 1\text{A}$ current will from $a$ to $b$

Why this question ? $\underset{–}{\text{Tip}}:$Whenever more than one battery is present in a closed circuit, always assume a current $i$ in the circuit and then apply $\text{KVL}$ in the closed loop.