What are Kirchhoff's two laws for the electrical circuit? Find out the reading of the ammeter with the help of the given circuit, while its resistance is negligible.
The sum of currents entering a junction is equal to the sum of currents leaving the junction.
\[ \sum I_{\text{in}} = \sum I_{\text{out}} \]
Kirchhoff's Second Law (KVL - Loop Rule):The sum of all voltages around any closed loop in a circuit is zero.
\[ \sum V = 0 \]
Solution for Ammeter Reading:Applying Kirchhoff's loop rule to the given circuit:
\[ \frac{5}{4 + 2} = \frac{5}{6} \, \text{A} \]
The voltage across the parallel resistors is:
\[ V = IR = \frac{5}{6} \times 2 = \frac{10}{6} \text{ V} \]
Current through the 6Ω resistor:\[ I = \frac{10}{6} \div 6 = \frac{5}{18} \, \text{A} \]
\( \text{Ammeter reading } = \frac{5}{18} \text{ A} \)
The switch (S) closes at \( t = 0 \) sec. The time, in sec, the capacitor takes to charge to 50 V is _________ (round off to one decimal place).
The op-amps in the following circuit are ideal. The voltage gain of the circuit is __________(round off to the nearest integer).
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The transformer connection given in the figure is part of a balanced 3-phase circuit where the phase sequence is “abc”. The primary to secondary turns ratio is 2:1. If \( I_a + I_b + I_c = 0 \), then the relationship between \( I_A \) and \( I_{ad} \) will be:
In the circuit shown below, if the values of \( R \) and \( C \) are very large, the form of the output voltage for a very high frequency square wave input is best represented by:
(b) Order of the differential equation: $ 5x^3 \frac{d^3y}{dx^3} - 3\left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^4 + y = 0 $