An RLRL circuit with with a 1212-ΩΩ resistor and a 0.06−H0.06−H inductor carries a current of 1A1A at t=0t=0, at which time a voltage source E(t)=12cos(120t)E(t)=12cos⁡(120t) VV is added. Determine the subsequent inductor current I(t)I(t). Hint: use the equation LdIdt+RI=E(t)LdIdt+RI=E(t) where L=L=inductance in henry R=R=resistance in ohm I=I=current in amp E=E=voltage in volts

The equation of current is

Explanation:

Resistance, R = 12 ohm

Inductance, L = 0.06 H

E (t) = 12 cos (120 t)

Compare with the standard equation,

So, the inductive reactance is

XL = w L = 120 x 0.06 = 7.2 ohm

The impedance of the circuit is

The current leads  by 90degree so the equation of current is

answer: strength i guess

explanation:

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz