What current will flow through the `2 K Omega` resistor in the circuit shown in the figure-

Options

A. 3 mA
B. 6 mA
C. 12 mA
D. 36 mA

Answer

A

Explanation

In the given circuit `4 kOmega` and `2 kOmega` are in series. This combination is in parallel with `3 Komega` resistance. Their effective resistance is-
`= ((4 + 2) xx 3)/((4 + 2) + 3) = 2K Omega`
This `2 KOmega` resistance is in series with `6 KOmega`. Thus, the equivalent resistance of the circuit is-
`R_(eq) = (2 KOmega) + (6 KOmega) = 8 KOmega)`
Current in the circuit is-
`I = (72V)/(8 xx 10^(3) Omega) = 9 xx 10^(-3) A = 9 mA`
Current through `2K Omega` resistor is-
`I_(1) = I xx 1/3 = (9 mA)/3 = 3 mA`