In the given circuit, if dI/dt = -1 A/s then the value of (VA - VB) at this instance will be (a) 30 V (b) 24 V (c) 18 V (d) 9 V

Answer: (a). Go from A to B along the current (2 A, flowing from A into the resistor). Drop across 12 Ω: iR = 24 V. Drop across inductor: LdI/dt = 6 × (-1) = -6 V (a rise of 6 V). Battery: going from its long (+) plate to its short (−) plate is a drop of 12 V. VA - VB = 24 - 6 + 12 = 30 V So the answer is (a) 30 V.

In the given circuit, if dI/dt = -1 A/s then the value of (VA - VB)… | MHT CET 2023 (12 May, Shift 1)

Q30MHT CET 2023 · 12 MAY · SHIFT 1Easy

Series LCR Circuit

In the given circuit, if $\frac{dI}{dt} = -1\text{ A/s}$ then the value of $(V_A - V_B)$ at this instance will be

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Answer: (a)

Go from A to B along the current (2 A, flowing from A into the resistor).
Drop across $12\ \Omega$ : $iR = 24$ V. Drop across inductor: $L\frac{dI}{dt} = 6 \times (-1) = -6$ V (a rise of 6 V).
Battery: going from its long (+) plate to its short (−) plate is a drop of 12 V.
$V_A - V_B = 24 - 6 + 12 = 30$ V
So the answer is (a) 30 V.

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Series LCR Circuit question from MHT CET 2023 (12 May, Shift 1)

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