Question

The network shown in figure is a part of the circuit. (The battery has negligible resistance.)At a certain instant the current I = 5 A and is decreasing at a rate of 10 ³ As⁻¹.What is the potential difference betweenpoints B and A?choose the correct option from the given options.
(A) 5 V
(B) 10 V
(C) 15 V
(D) 0 V

Answer 1

see figure given,

Resistance, R = 1 Ω, emf , ξ = 15V and inductance , L = 5mH, di/dt = 10³ A/s

so, $V_A-IR+\mathfrak{E}-L\frac{di}{dt}-V_B=0$

⇒ $V_A$ - 5 × 1 + 15 - 5 × 10⁻³ × 10³ - $V_B$ = 0

⇒ $V_A$ -$V_B$ - 5 + 15 - 5 = 0

⇒$V_A$ -$V_B$ + 15 = 0

⇒ $V_B$ -$V_A$  = 15V

hence option (c) is correct.

Answer 2

Answer:

option:(c) 15 V

Explanation:

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