For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals ‘ab’ is

The correct option is: 65 V in series with 15 Ω.

To find the equivalent Thevenin voltage ($V_{\mathrm{th}}$) and Thevenin impedance ($R_{\mathrm{th}}$) across terminals ab, we analyze the circuit in two main steps:

Step 1: Find the Thevenin Voltage ($V_{\mathrm{th}}$)
Under open-circuit conditions across terminals ab, no current can flow through the branch containing the $2\Omega$ resistor and the dependent voltage source $3i_1$.
Consequently, the entire current from the independent current source of $5A$ flows down through the middle $10\Omega$ resistor. Since $i_1$ is defined as the downward current in this branch, we have:
$i_1=5A$
The voltage drop across the $10\Omega$ resistor (which is the node voltage above it, relative to terminal b) is:
$V_{10}=i_1\times 10=5A\times 10\Omega =50V$
Applying Kirchhoff's Voltage Law (KVL) to find the open-circuit voltage at terminals ab:
$V_{\mathrm{th}}=V_{\mathrm{ab}}=V_{10}+3i_1$
Substituting the values of $V_{10}$ and $i_1$:
$V_{\mathrm{th}}=50V+3(5A)=50+15=65V$

Step 2: Find the Thevenin Impedance ($R_{\mathrm{th}}$)
To determine the equivalent resistance, we deactivate the independent sources. The $5A$ independent current source is replaced with an open circuit. Next, we apply an external test voltage source $V_x$ across terminals ab that drives a test current $I_x$ into terminal a.
With the left-hand source branch open, the current $I_x$ passes through the $2\Omega$ resistor, the dependent source, and continues downwards through the $10\Omega$ resistor to terminal b. Consequently, the current $i_1$ is identical to $I_x$:
$i_1=I_x$
Applying KVL from terminal a to terminal b:
$V_x=(I_x\times 2)+3i_1+(I_x\times 10)$
Substituting $i_1=I_x$:
$V_x=2I_x+3I_x+10I_x$
$V_x=15I_x$
Solving for the equivalent resistance:
$R_{\mathrm{th}}=\frac{V_x}{I_x}=15\Omega$

Therefore, the equivalent Thevenin network consists of a 65 V voltage source in series with a 15 Ω resistor.