AB is a part of an electrical circuit (see figure). The potential difference $V_{A}-V_{B}$ at the instant when current $i=2~A$ and is increasing at a rate of 1 amp/second is:

Correct Answer: 10 volt

Step I: Write the KVL equation starting from point A to point B: $V_A - iR - L\frac{di}{dt} - V = V_B$.
Step II: Identify the given values: current $i = 2~A$, rate of change of current $\frac{di}{dt} = 1~A/s$, resistance $R = 2~\Omega$, inductance $L = 1~H$, battery voltage $V = 5~V$.
Step III: Substitute the values into the equation: $V_A - V_B = iR + L\frac{di}{dt} + V$.
Step IV: Calculate the potential difference: $V_A - V_B = (2 \times 2) + (1 \times 1) + 5 = 4 + 1 + 5 = 10~V$.

सही उत्तर: 10 volt

Step I: बिंदु A से बिंदु B तक KVL समीकरण लिखें: $V_A - iR - L\frac{di}{dt} - V = V_B$।
Step II: दिए गए मानों को पहचानें: धारा $i = 2~A$, धारा के परिवर्तन की दर $\frac{di}{dt} = 1~A/s$, प्रतिरोध $R = 2~\Omega$, अधिष्ठापन (inductance) $L = 1~H$, बैटरी वोल्टेज $V = 5~V$।
Step III: मानों को समीकरण में प्रतिस्थापित करें: $V_A - V_B = iR + L\frac{di}{dt} + V$।
Step IV: संभावित अंतर की गणना करें: $V_A - V_B = (2 \times 2) + (1 \times 1) + 5 = 4 + 1 + 5 = 10~V$।