Question

Hello, can i get help solving this question in physics 2 ?

uestion3 A 12V battery is applied to the terminals of a resistor of 1KSQ resistance a) Find the current in the circuit. b) Find the power dissipated in the resistor. c) Assume that the resistor is removed from the circuit, and two in-series capacitors are put in its CL place, where one capacitor has a capacitance of 25nF and the other has a capacitance of 35nF Find the energy stored in each capacitor.

Answer 1

For finding current in the circuit, we will use ohm's law. V = IR V = 12v, R = 1 k ohm therefore 12 = I times 1000 I = 12/1000 = 12 times 10^-3 I = 12 Ma b) Power dissipiated in the resistor P = VI = 12 times 12 times 10^-3 = 144 times 10^-3 = 0.144 watt \r\n c) Since both the capacitor are in series, so they will get same charge. Now charge flouring in the circuit can be calculated as. q = C_net V 1/C_net = 1/C_1 + 1/ C_2 = C_1 + C_2/ C_1 C_2 C_net = C_1 C_2/ C_1 + C_2 = (25 times 35/25 + 35) times (10^-9)^2/10^-9 = 875/60 times (10^-9)^2/(10^-9) = 875/60 nF C_net = 14.58 nF Now q = 875/60 nF times 12 v q = 175 Nc \r\n Now Energy Calculation First we will find potential across each capacitor. potential