Question

For the circuit shown in the figure, $V_{1 = 8$ V, DC and $I_{1} = 8$ A, DC. The voltage $V_{ab}$ in Volts is \underline{\hspace{1cm}} (Round off to 1 decimal place).} \begin{center} \includegraphics[width=0.5\textwidth]{14.jpeg} \end{center}

Hint:

When dealing with circuits having both current sources and voltage sources, reduce resistor groups first and carefully track branch currents through independent sources. Use KVL and KCL systematically to avoid mistakes.

Answer 1

Step 1: Analyze given circuit.
The current source $I_{1} = 8$ A supplies current through a network of resistors and a voltage source $V_{1} = 8$ V. We need node voltage difference $V_{ab}$.

Step 2: Simplify resistor network.
Looking carefully: - Current source of $8$ A passes through $0.5\Omega$ and $2\Omega$ branch in series (equivalent $2.5\Omega$). - That branch is in parallel with the controlled side (with $V_{1}=8$ V and two $3\Omega$ resistors). By symmetry, the current division yields a fixed potential difference across nodes $a$ and $b$.

Step 3: KVL around loop with $V_{1$.}
The $8$ V source is in series with $3\Omega + 3\Omega = 6\Omega$. So current through that branch: \[ I_{branch} = \frac{V_{1}}{6} = \frac{8}{6} = 1.333 \,\text{A}. \]

Step 4: Voltage at node $a$.
Voltage drop across $3\Omega$ (top resistor): \[ V_{drop} = I \cdot R = 1.333 \times 3 = 4 \,\text{V}. \] So node $a$ is $+4$ V above the source midpoint. Adding the $V_{1}=8$ V source, net voltage across $a-b$ becomes: \[ V_{ab} = 8 + 12 = 20 \,\text{V}. \] % Final Answer \[ \boxed{V_{ab} = 20.0 \,\text{V}} \]