Test: Basic Laws - 1 - Electronics and Communication Engineering (ECE) MCQ

On combining the resitances in parallel, equivalent circuit will reduce as shown in below.

Hence, equivalent resistance is

The equivalent resistance of the given circuit is

Hence, for current I to be maximum , R should be minimum.

Converting the current sources into equivalent voltage sources, the circuit is reduced as shown below.

Equivalent voltage in the given circuit is

V= 6 + 2 0 - 1 0 = 16 volt

And equivalent resistance is R = 1 + 5 + 2 = 8 Ω

Converting all the voltage sources into equivalent current sources using source transformation, the equivalent circuit will be reudced as shown below.

Total power drawn from the circuit = 2 x 100 = 200 watts

Hence, supply current is

Let R be the series resistance to be inserted in the circuit such that the voltage across the bulbs is 220 V.

Applying KVL , we get

Converting the star-connected resistors into Δ equivalent, the given circuit is reduced as shown below.

Hence, equivalent resistance between terminals Xand Y is

KCL is based on law of conservation of charge while KVL is based on law of conservation of energy.

The current through the resistors will purely depend upon the voltage source of 2V connected a cross them . Let the currents i₁ , i₂  and i₃ respectively flowing through the resistors of 1Ω,  2Ω and 5Ω.