Question

Not yet answered Points out of 6.00 P Flag question In the circuit below R1 = 7.51, R2 = R3=5, and L = 31 mH. The battery supplies 12 V. R1 R3 8 R2 ell (b) What is the current through the battery after the switch has been closed for a long time? Select one: O a. 0.00 A O b. 1.20 A O c. 1.60 A O d. 0.53 A O e. 0.96 A

Answer 1

After the swtich has been closed for a long time , the inductor acts as merely a connecting wire (for DC connections).

Thus, the resistance R2 and R3 are in parallel.

So, R_eq= R3*R2/(R3+R2).

and thus, total resistance

$R_{tot}= R_{1} + \frac{R_{3}R_{2}}{R_{3}+R_{2}}$

for R1=7.5Ω and R2=R3=5Ω

we get Rtot = 10Ω

So, Current across battery = V/R=12/10 = 1.2A

Option B is correct