Question

In the circuit shown below, the emf of the battery E=42 volts; the resistors have values R1 = 116 12, R2 = 184 12, and R3 = 139 N; and the capacitor has a capacitance C = 5 UF (microfarads). After the switch has been closed for a long time, what is the current supplied to the circuit by the battery? Answer in milliamps to the nearest milliamp, where 1000 milliamps = 1 amp. S R} E I RA С w R3

Answer 1

as are E R2 w O 0 After closing switch so circuit can be redrawn follows: Where resistances of resistors Ri Ri=116 r. R2=1842 С R3 =) 39 2 (Fig) Capacitance of capacitor is C = MF EMF of bat trrey is E = 42 V When switch then S is closed for Capacitor charged fully and acts an open circuit and Current flows through Capacitor. Now, (Fig 1) can be redrawn follows R3 long time no RI W Rq OPEN CIRCUIT (Fig 2) www R3 As no current flows in in open circuit ,so (Fig2) be redrawn can as follows I I RI + E + R2 (Fig3) fum II I R3 From (Fig 3), we see that resistors Ri, Rz and R3 are in series same current flow through each resistor. Let current is I. This aurrent I is supplied by battery. Now, use Kirchoff's Voltage Law (KUL) in loop of , so this

2 :) (Fig 3), me get (KVL says that the sum of potential drop's across each elem ent in a loop is zeno Potential drop across resistor = current X resistance Now, we hare E - IR, - I R2 -I R2-IR3 E I CRIT R2 + R3) -) E = ICR, + R2+R3) 5) I = E 42 V =) I = R1 + R2TR3 1162 +184 +1392 42 V I = > I = 0.095672 A [:11 I 439 m [:** = A 60 I A = 10oom =) I 0.095672X1000mA So, I = 95.672 m ñ 9.6 mA current supplied to the circuit by the battery 96 m is