Question

For the circuit shown below :
The current $i_0$ is

• A. -3.2 mA
• B. - 4 mA
• C. - 4.8 mA
• D. -0.8 mA

Answer 1

The correct option is C - 4.8 mA


This is a summmer with two inputs,

$V_0=-\left[\frac{10}{2}\right(2)+\frac{10}{2.5}(1\left)\right]$
= - (4+4)
= - 8 V

The current $i_0$ is the sum of the current through the $10k\Omega$ and $2k\Omega$ resistors. Both the resistors have voltage $V_0=-8V$ across them ,
Since, $V_a=V_b=0$

Hence, $i_0=\frac{V_0-0}{10}+\frac{V_0-0}{2}$
= - 0.8 - 4 = - 4.8 mA