Circuits

Parallel Resistors Checkpoint Two resistors of very different value are connected in parallel. Will the resistance of the pair be closer to the value of the larger resistor or the smaller one? – the larger resistor – the smaller resistor

checkpoint Car Headlights Are car headlights connected in series or parallel? Parallel

As more identical resistors R are added to the parallel circuit shown here, the total resistance between points P and Q 1.increases 2.remains the same 3.decreases Q

As more identical resistors R are added to the parallel circuit shown here, the total resistance between points P and Q 1.increases 2.remains the same 3.decreases Q

Charge flows through a light bulb. Suppose a wire is connected across the bulb as shown. When the wire is connected, 1.all the charge continues to flow through the bulb. 2.half the charge flows through the wire; the other half continues through the bulb. 3.all the charge flows through the wire. 4.None of the above

Charge flows through a light bulb. Suppose a wire is connected across the bulb as shown. When the wire is connected, 1.all the charge continues to flow through the bulb. 2.half the charge flows through the wire; the other half continues through the bulb. 3.all the charge flows through the wire. 4.None of the above

Power Power is the rate at which energy is used or at which work is done P = IV Units:

Practice: Resistors in Series Calculate the power provided by the battery if the battery emf is 22 volts. Calculate the power dissipated by each resistor R 12 = R 1 + R 2 I 12 = V/R 12 P = IV = 11  R 12 00 = I 12 = 2 Amps = 2 A*22 V = 44 W Expand: V 1 = I 1 R 1 P = IV V 2 = I 2 R 2 P = IV = 2 x 1 = 2 Volts =2 A * 2 V = 4 W = 2 x 10 = 20 Volts = 2 A * 20 V = 40 W R 1 =1  00 R 2 =10  Check: P 1 + P 2 = P battery ? Simplify (R 1 and R 2 in series): R 1 =1  00 R 2 =10 

Practice: Resistors in Parallel Determine the current through the battery. Let  = 60 Volts, R 2 = 20  and R 3 =30 . 1/R 23 = 1/R 2 + 1/R 3 V 23 = V 2 = V 3 I 23 = I 2 + I 3 R2R2 R3R3  R 23  R 23 = 12  = 60 Volts = V 23 /R 23 = 5 Amps Simplify: R 2 and R 3 are in parallel

Practice: Resistors in Parallel What is the power delivered by the battery and what is the power dissipated by each resistor. Let  = 60 Volts, R 2 = 20  and R 3 =30 . P = I*V P 2 = I 2 V 2 P 3 = I 3 V 3 R2R2 R3R3  R 23  = (5 A)(60 V) = 300 W = (3 A)(60 V) = 180 W = (2 A)(60V) = 120 W Calculate IV for the battery.

Try it! R1R1 R2R2 R3R3 Calculate current through each resistor. R 1 = 10 , R 2 = 20  R 3 = 30    V Simplify: R 2 and R 3 are in parallel 1/R 23 = 1/R 2 + 1/R 3 V 23 = V 2 = V 3 I 23 = I 2 + I 3 Simplify: R 1 and R 23 are in series R 123 = R 1 + R 23 V 123 = V 1 + V 23 =  I 123 = I 1 = I 23 = I battery : R 23 = 12  : R 123 = 22  R 123  R1R1 R 23   : I 123 = 44 V/22  A Power delivered by battery? P=IV = 2  44 = 88W

Try it! (cont.) R1R1 R2R2 R3R3 Calculate current through each resistor. R 1 = 10 , R 2 = 20  R 3 = 30    V Expand: R 2 and R 3 are in parallel 1/R 23 = 1/R 2 + 1/R 3 V 23 = V 2 = V 3 I 23 = I 2 + I 3 Expand: R 1 and R 23 are in series R 123 = R 1 + R 23 V 123 = V 1 + V 23 =  I 123 = I 1 = I 23 = I battery R 123  R1R1 R 23   : I 23 = 2 A : V 23 = I 23 R 23 = 24 V I 2 = V 2 /R 2 =24/20=1.2A I 3 = V 3 /R 3 =24/30=0.8A

Checkpoint 4 Resistor Combination Is it possible to connect 4 resistors of resistance R in such a way that their equivalent resistance is R?

If the 4 light bulbs in the figure are identical, which circuit puts out more light? 1.I 2.They emit the same amount of light 3.II I II

If the 4 light bulbs in the figure are identical, which circuit puts out more light? 1.I 2.They emit the same amount of light 3.II I II