Question

The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is 5A and it is decreasing at a rate of ${10}^{3}A{s}^{–1}$ then $V_B–V_A$ equals

• A. $20\text{V}$
• B. $15\text{V}$
• C. $10\text{V}$
• D. $5\text{V}$

Answer 1

The correct option is B $15\text{V}$

$V_A-V_B=IR-15-L\frac{di}{dt}$
$V_A-V_B=5\times 1-15-5\times {10}^{-3}\times {10}^3$
(since the current is decreasing, the polarity of induced e.m.f is such that it intends to increase the current)
$V_A-V_B=-15V$
$V_B-V_A=15V$