 Tell the electric company how much electrical energy is used  Measure kW hrs (energy units like Joules)  In most homes and factories outlets vary.

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Presentation transcript:

 Tell the electric company how much electrical energy is used  Measure kW hrs (energy units like Joules)  In most homes and factories outlets vary in voltage (110 V, 220 V, etc)  To determine energy use per outlet  V=IR, I = q/t, W = Vq  Meters measure voltage and current instantaneously and calculate power and energy

Power = energy = Vq time t Since I = q t Power, P = VI Units: (A)(V) = C J = J = W s C s **Electrical meters measure kW hrs → (P)(t)

Devices (TV, Refrigerator, etc) are rated for maximum power/voltage/etc. (They stop working if exceeded) Devices can be overpowered by accident by  overloading outlets – draws too much current  Example: plugging too many things into the same outlet  Ground Fault (short circuit) – contact is made between live and ground conductors  Example: Working hair dryer in water – plumbing is connected to ground

Preventing Accidental Overload  Fuses  small filaments (like light bulbs)  Burn out when too much current is drawn  Circuit breakers  bimetallic strip heats up when too much current is drawn  bends when heated which opens switch to prevent current flow  GFI outlets and breakers  Complex circuitry that senses too much current almost instantly  Opens circuit at outlet or breaker

0.00l A and higher – feel shock 0.01A and higher – unable to release 0.02 A and higher – paralyzes respiratory muscles (you can’t breath) 0.1 A and higher – ventricular fibrillation (erratic heartbeat) 1.0 A and higher – “cooked”

Skin has very high resistance – approximately 100,000 Ω So, V = IR 10 V = I (100,000 Ω ) I = A Remember! You can’t have a current without a voltage

To calculate the power dissipated in a resistor: P = IV Or If I is unknown then because V = IR, P = V 2 /R If V is unknown then because V = IR, P = I 2 R To calculate the total power dissipated by the circuit use the total current and total voltage

Total power dissipated in the circuit = sum of the power dissipated in each resistor P total = P 1 + P 2 Figure 3 R1R1 R2R2 R1R1 R2R2