For the difference amplifier circuit shown, determine the output voltage at terminal $A$.
A. $ - 18.13\;{\rm{V}}$.
B. $ - 6.07\;{\rm{V}}$.
C. $6.07\;{\rm{V}}$.
D. $15.45\;{\rm{V}}$.

Hint:In this solution we will take the equivalent resistance when the resistances are connected in the parallel or series connection and will take the voltages of voltmeters to find out the output voltage at the given terminal.

Complete step by step answer:
Voltage ${V_1} = 30\;{\rm{V}}$.
Voltage ${V_2} = 25\;{\rm{V}}$.
Resistance ${R_1} = 15\;\Omega $.
Resistance ${R_2} = 20\;\Omega $.
Resistance ${R_3} = 5\;\Omega $.
Resistance ${R_4} = 3\;\Omega $.
When a circuit or device has a power gain more than one then that circuit is known as an amplifier. The amplifiers are mostly used in the electronic components.
Express the formula of the output voltage of the difference amplifier.
${V_A} = - \dfrac{{{R_2}}}{{{R_1}}}{V_1} + \left( {\dfrac{{{R_4}}}{{{R_3} + {R_4}}}} \right)\left( {1 + \dfrac{{{R_2}}}{{{R_1}}}} \right){V_2}$
Here,
${V_A}$ is the voltage at $A$.
${R_2}$,${R_1}$,${R_3}$ and ${R_4}$ are the resistance.
$V{}_1$ and ${V_2}$ are the voltage.
Substitute $30\;{\rm{V}}$ for ${V_1}$ , $25\;{\rm{V}}$ for ${V_2}$ , $15\;\Omega $ for ${R_1}$ , $20\;\Omega $ for ${R_2}$, $5\;\Omega $ for ${R_3}$ and $3\;\Omega $ for ${R_4}$ to obtain the voltage at $A$.
$
{V_A} = - \dfrac{{20\;\Omega }}{{15\;\Omega }}\left( {30\;{\rm{V}}} \right) + \left( {\dfrac{{3\;\Omega }}{{3\;\Omega + 5\;\Omega }}} \right)\left( {1 + \dfrac{{20\;\Omega }}{{15\;\Omega }}} \right)\left( {25\;{\rm{V}}} \right)\\
\Rightarrow{V_A} = - 40 + \dfrac{{21}}{{25}}\left( {25\;{\rm{V}}} \right)\\
\Rightarrow{V_A} = - 40 + \dfrac{{525}}{{24}}\;\;{\rm{V}}\\
\Rightarrow{V_A} = - 40 + 21.875\;{\rm{V}}\\
\Rightarrow {V_A} = - 18.125\;{\rm{V}}\\
\therefore{V_A} = - 18.13\;{\rm{V}}
$
Therefore, the correct option is (A).

Note:While solving this type of questions make sure that the prediction of parallel or series connection must be right to find out the equivalent resistance otherwise the solution will be wrong.