Question

Plz answer this quickly. I will mark u brainliest.

Answer 1

Answer: 0A

Explanation:

Kirchhoff’s rules are as follows:

(i) Junction rule (or current rule): The sum of all currents entering a junction is equal to sum of all currents leaving the junction.

(ii) Voltage rule: The algebraic sum of changes in the potential around any closed loop must be zero.

Apply Kirchhoff’s voltage rule 2 in loop ABCDA: Apply Kirchhoff’s voltage rule 2 in loop FEDCF:

Solving Eqs. (1) and (2) [and multiplying Eq. (1) by 4 and Eq. (2) by 3 and add], we get Substituting i2 = 4 in Eq. (1), we get Therefore, the current passing through 20 Ω resistor is 4 A and the current passing through 40 Ω resistor is i1 – i2 = 4 A – 4 A = 0 t