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Q. The network shown in figure is a part of a complete circuit. If at a certain instant the current $'I'$ is $5A$ and is decreasing at the rate of $10^{3}As^{- 1}$ , then $V_{A}-V_{B}$ is :
Question

NTA AbhyasNTA Abhyas 2022

Solution:

$\frac{d i}{d t}=-10^{3}A/sec$
Solution
Apply KVL
$V_A-5 \times 1-15 V-5 \times 10^{-3} \times\left(\frac{d i}{d t}\right)-V_B=0$
$V_A-5 \times 1-15-5 \times 10^{-3} \times\left(-10^3\right)-V_B=0 $
$V_A-V_B=15 V$