The voltage across a 5Ω resistor is given as V = 200 sin(100πt). Find out the time required for current through io it to change from\(\frac{i_0}{2}\) to i0 [ i0 is peak current]
\(\frac{1}{300}\) s
\(\frac{1}{600}\) s
\(\frac{1}{150}\) s
\(\frac{1}{1200}\) s
The current passing through the battery in the given circuit, is:
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
Displacement current is a quantity appearing in Maxwell’s equations. Displacement current definition is defined in terms of the rate of change of the electric displacement field (D). It can be explained by the phenomenon observed in a capacitor.