The output of a automobile alternator is typically about 14V D. C. The headlight circuit draws about 4 Amps of current. Using Ohm's Law (V = IR), the resistance of the headlight bulb can be calculated. Suppose your alternator quits working as you are driving home. The battery voltage now supplying energy to the headlight bulb drops to 12V D. C. Assume the bulb resistance R remains constant (it won’t, but that is another story). Required:
How much current (in Amps) now flows through the bulb?

I = 3.42 A

Explanation:

Ohm's law gives a relationship between the potential difference and the current passing through a conductor. The formula of Ohm's Law is given as follows:

V = IR

where,

V = Voltage

I = Current

R = Resistance

Initially:

V = 14 volts

I = 4 A

R = ?

Therefore,

14 volts = (4 A)R

R = (14 volts)/(4 A)

R = 3.5 Ω

Now, when V becomes 12 volts and R remains constant:

V = IR

12 volts = I(3.5 Ω)

I = (12 volts)/(3.5 Ω)

I = 3.42 A

  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