Question

In the circuit shown below, the current that flows from a to b when the switch S is closed is

• A. + 1.0 A
• B. + 1.5 A
• C. - 1.0 A
• D. - 1.5 A

Answer 1

The correct option is C - 1.0 A


When S is closed $5\Omega$ will be useless because of short-circuit
So, current will pass from $10\Omega and20\Omega$

$\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}=\frac{1}{10\Omega }+\frac{1}{20\Omega }$

$r_{eq}=\frac{20}{3}\Omega$

$I=\frac{10W}{\frac{20}{3}}=\frac{3}{2}A$

Now,
$I=\frac{20}{10+20}I=\frac{20}{10+20}\times \frac{3}{2}$

$I_1=1A\to$ this is from b to a.

So current from a to b = -I = -1 A

Hence current from a to b = -1

Hence, option (d) is correct