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Potentiometer

In the branch...

Question

In the branch AB of a circuit, as shown in the figure, a current $I = (10 t + 5) A$ is flowing, where t is the time in second. At t = 0, the value of $(V_{A} - V_{B})$ in Volt.

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Solution

The correct option is B 15
$V_{A} - V_{B} = ε + L \frac{d i}{d t} + R$
Where $ε = 10 V$
$\frac{d i}{d t} = 10 A / s$ at $t = 0$
i = 5A at t = 0

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Similar questions

A B

I = (10 t + 5) A

t

t = 0

A

B

(V_{A} - V_{B})

i = 2 t - 8

t = 1 s

V_{a} - V_{b} = + 4 V

V_{a} - V_{b}

+ 4 V

i = 2 t - 8

t = 1 s

V_{a} - V_{b} = + 4 V

V_{a} - V_{b}

+ 4 V

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5 A

10^{3} {As}^{- 1}

V_{B} - V_{A}

(i)

(t)

i = 2 t^{2} + 5

t

second

t = 1 s

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