Question

In the circuit shown below, the current $i$ flowing through $200\,\Omega$ resistor is ________ mA (rounded off to two decimal places).

Hint:

A series (voltage source $V_s$ + $R$) to ground can be replaced by a Norton source of $I_N=\dfrac{V_s}{R}$ from ground to the node, in parallel with $R$. This often makes nodal analysis straightforward.

Answer 1

Take the bottom node as reference (0 V). Let the left top node be $V_A$ and the right top node (top of $200\,\Omega$) be $V_B$.
Replace the left branch (a $2\,$V source in series with $2\,\text{k}\Omega$ to ground) by its Norton equivalent: a $1\,$mA source from ground to node $A$ in parallel with $2\,\text{k}\Omega$.
This is in parallel with the vertical $1\,$mA source (ground $\to A$) and another $2\,\text{k}\Omega$ to ground, giving net: two $2\,\text{k}\Omega$ in parallel $\Rightarrow 1\,\text{k}\Omega$ from $A$ to ground, and total current injection at $A$ of $2\,$mA.
Between $A$ and $B$ there is a $1\,\text{k}\Omega$ resistor in parallel with a $1\,$mA source from $A$ to $B$. The load is $200\,\Omega$ from $B$ to ground.
Nodal equations (currents injected taken positive):
At $A$: $\displaystyle \frac{V_A}{1\,\text{k}\Omega}+\frac{V_A-V_B}{1\,\text{k}\Omega}=2\,\text{mA}-1\,\text{mA}=1\,\text{mA}$.
At $B$: $\displaystyle \frac{V_B-V_A}{1\,\text{k}\Omega}+\frac{V_B}{200\,\Omega}=+1\,\text{mA}$.
Solving, $V_A=0.6364\,$V and $V_B=0.2727\,$V.
Current through $200\,\Omega$ (downward) is $i=\dfrac{V_B}{200}=0.0013636\,$A $=1.36\,$mA (to two decimals).
\[ \boxed{i \approx 1.36~\text{mA}} \]