MHT CET · Physics · Electromagnetic Induction
The network shown in the figure is part of complete circuit. If at a certain instant the current \(i\) is \(5 \mathrm{~A}\) and is decreasing at the rate of \(10^3 \mathrm{~A} / \mathrm{s}\), then \(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}\) is
- A 15V
- B 10V
- C 5V
- D 20V
Answer & Solution
Correct Answer
(A) 15V
Step-by-step Solution
Detailed explanation
Consider the diagram below:
Given, \(=5 \mathrm{~A}, \mathrm{R}=1 \Omega, \mathrm{L}=5 \mathrm{mH}, \frac{\mathrm{di}}{\mathrm{dt}}=10^3 \mathrm{~A} / \mathrm{s}\)
\(\begin{aligned}
& V_A-i R+E-L\left(\frac{d i}{d t}\right)=V_B \\
& \Rightarrow V_A-V_B=\left(i R+L\left(\frac{d i}{d t}\right)-E\right)
\end{aligned}\)
\(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=5 \mathrm{~A} \times 1 \Omega+5 \times 10^{-3} \mathrm{H} \times 10^{-3}\) \(\frac{\mathrm{A}}{\mathrm{s}}-15 \mathrm{~V}\)
\(\mathrm{~V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=-(5+5-15) \mathrm{V}=5 \mathrm{~V}\)
Given, \(=5 \mathrm{~A}, \mathrm{R}=1 \Omega, \mathrm{L}=5 \mathrm{mH}, \frac{\mathrm{di}}{\mathrm{dt}}=10^3 \mathrm{~A} / \mathrm{s}\)
\(\begin{aligned}
& V_A-i R+E-L\left(\frac{d i}{d t}\right)=V_B \\
& \Rightarrow V_A-V_B=\left(i R+L\left(\frac{d i}{d t}\right)-E\right)
\end{aligned}\)
\(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=5 \mathrm{~A} \times 1 \Omega+5 \times 10^{-3} \mathrm{H} \times 10^{-3}\) \(\frac{\mathrm{A}}{\mathrm{s}}-15 \mathrm{~V}\)
\(\mathrm{~V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=-(5+5-15) \mathrm{V}=5 \mathrm{~V}\)
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