Parallel Circuit- The voltages across each resistor is the same with the voltage source:

VT = V1 = V2 = V3 = … = VN - The total current (IT) flowing through the circuit is equal to the sum of current flowing each resistor:

IT = I1 + I2 + I3 + … + IN - The total resistance is given by:

1 2 3

1 1 1 1 1...

T NR R R R R

IT

I1

I2

I3

Sample problem

= 2.20 kΩ

= 1.00 kΩ

= 10.0 kΩ

Find the total current and total resistance of the circuit. Find also the individual voltages across each resistor and the current flowing each resistor.

= 8.54 V

Given: R1 = 2.20 kΩ R2 = 1.00 kΩ R3 = 10.0 kΩ

VT = 8.54 V

Solutions:The total resistance, RT is given by

1 2 3

1 1 1 1

1 1 1 1

2.20 1.00 10.0

1 10.0 22.0 2.20 34.2

22.0 22.0

22.00.643 643

34.2

T

T

T

T

R R R R

R k k k

R k k

kR k

Note that the total resistance is less than the smallest resistance of the resistors connected in parallel and that is always the case for this type of circuit.

Solving the total current, IT

IT = VT/RT = 8.54V/643Ω = 0.0133 A

Since the resistors are connected in parallel, VT = V1 = V2 = V3 = 8.54 V

For the individual currents:I1 = V1/R1 = 8.54 V/2.20x103 Ω = 0.00388 A

I2 = V2/R2 = 8.54 V/1.00x103 Ω = 0.00854 A

I3 = V3/R3 = 8.54 V/10.0x103 Ω = 0.000854 A

Actual Circuit

R1 = 2.20 kΩR2 = 1.00 kΩ

R3 = 10.0 kΩ

Electric Power

- Is the rate at which electrical energy is converted to other forms of energy (e.g. light, heat, mechanical, etc.).

P = VI = I2R = V2/Rwhere:

V = voltage in Volts (V)I = current in Amperes (A)P = electric power in Watts (W)R = resistance in Ohms (Ω)

Electrical EnergyEE = Pt

where:P = electrical power in kiloWatts (kW)t = time in hours (h)EE = electrical energy consumption in

kiloWatthour (kWh)

EE cost = (EE) x (cost/kWh)

Sample problem

IT

= 8.54 V

= 2.20 kΩ = 1.00 kΩ = 10.0 kΩ

Find the total power and the individual power in each resistor in this series circuit.

= 0.000647 A

V1 = 1.42 V V2 = 0.647 V V3 = 6.47 V

Solutions

Given: R1 = 2.20 kΩ R2 = 1.00 kΩ R3 = 10.0 kΩ

VT = 8.54 V IT = I1 = I2 = I3 = 0.000647 A V1 = 1.42 V V2 = 0.647 V V3 = 6.47 V

Solution:P1 = V1I1 = (1.42 V)(0.000647A) = 0.000919 WP2 = V2I2 = (0.647 V)(0.000647A) = 0.000419 WP3 = V3I3 = (6.47 V)(0.000647A) = 0.00419 WPT = VTIT = (8.54 V)(0.000647A) = 0.00553 W

What current is drawn by a 50.0-W bulb plug in a 220 V source? If it is used for 6.00 hours, how much electrical energy is used? How much does it cost if VECO charges Php 7.50/kWh?

Given: P = 50.0 W or 0.0500 kW V = 220 V t = 6.00 h EE cost = Php 7.50/kWh

Solution:I = P/V = 50.0 W/220 V = 0.23 A (current drawn)

The electric energy consumption, EE is given byEE = Pt = (0.0500kW)(6.00 h) EE = 3.00 kWh

EE cost = EE x cost/kWh = (3.00 kWh)(Php 7.50/kWh)

EE cost = Php 22.5