Question

In figure, take, ${\epsilon }_1=6.0\text{V},{\epsilon }_2=1.5\text{V}$ and ${\epsilon }_3=4.5\text{V},R_1=600\Omega ,R_2=125\Omega$ and $R_3=300\Omega$. The correct option(s) is/are -

• A. Current in cell ${\epsilon }_1$ is away from its positive terminal
• B. Current in ${\epsilon }_2$ is is away from its negative terminal.
• C. Current in ${\epsilon }_3$ is away from its negative terminal.
• D. Current in ${\epsilon }_1$ is is away from its negative terminal.

Answer 1

Answer: (A), (B)

Current in ${\epsilon }_2$ is is away from its negative terminal.


In the left loop, on going clockwise from $P$, we get,

$-I{R}_1-(I+i)R_2-{\epsilon }_2+{\epsilon }_1=0$

$\Rightarrow -600I-125(I+i)-1.5+6=0$

$\Rightarrow 725I+125i=4.5...\left(1\right)$

Similarly, in the right loop, on going clockwise from $Q$, we get,

$-{\epsilon }_3+{\epsilon }_2+(I+i)R_2+i{R}_3=0$

$\Rightarrow -4.5+1.5+125(I+i)+300i=0$

$\Rightarrow 125I+425i=3...\left(2\right)$

From $1$ and $2$, we get,

$I=+ve$ and $i=+ve$

Hence, current in ${\epsilon }_1$ is away from its positive terminal, current in ${\epsilon }_2$ is away from its negative terminal. and current in ${\epsilon }_3$ is away from its positive terminal.

Therefore, options $\left(A\right)$ and $\left(B\right)$ are correct.