Question

In the branch AB of the circuit, as shown in the figure, a current $I = \left( {t + 2} \right)A$ is flowing, where $t$ is the time in second. At $t = 0$, the value of $\left( {{V_A} - {V_B}} \right)$is $x$, if $x$ is positive write value of $x$, if $x$ is negative, then write $\left| x \right| + 1$.

Answer 1

Hint:It is clearly seen that the given circuit is RL circuit. Therefore, to solve this question, we need to use Kirchhoff's voltage law for this RL circuit. For this first, we need to find the derivative of the current with respect to time and then we will apply Kirchhoff's law to determine the asked value.

Complete step by step answer:
We will first consider the given circuit and form the equation applying the Kirchhoff’s law.
${V_A} - IR - L\dfrac{{dI}}{{dt}} + 10 - {V_B} = 0$
For this we will first find the value of current $I$when time $t = 0$.
$I = \left( {t + 2} \right) = 0 + 2 = 2A$
Now, we will find the value of $\dfrac{{dI}}{{dt}}$
$
I = \left( {t + 2} \right) \\
\Rightarrow \dfrac{{dI}}{{dt}} = 1 \\ $
Now we will use the values ${V_A} - {V_B} = x$, $I = 2A$, $R = 3\Omega $,$L = 1H$ and $\dfrac{{dI}}{{dt}} = 1$in the equation:
\[
{V_A} - IR - L\dfrac{{dI}}{{dt}} + 10 - {V_B} = 0 \\
\Rightarrow x - \left( {2 \times 3} \right) - \left( {1 \times 1} \right) + 10 = 0 \\
\Rightarrow x + 3 = 0 \\
\Rightarrow x = - 3 \\ \]
Here, we are getting the negative value of $x$.
We are asked to write $\left| x \right| + 1$ if $x$ is negative.
$\therefore\left| x \right| + 1 = \left| { - 3} \right| + 1 = 3 + 1 = 4$

Thus, our final answer is 4.

Note:We have used the second law of Kirchhoff to solve this question. This law states that the voltage around a loop equals the sum of every voltage drop in the same loop for any closed network and also equals to zero. This means that the algebraic sum of every voltage in the loop has to be equal to zero and this property of Kirchhoff’s law is called conservation of energy.