Example: two 100  light bulbs are connected (a) in series and (b) in parallel to a 24 V battery. For which circuit will the bulbs be brighter? parallel (left) series (right)

(a) Series combination. R1 100 R2 100 Req = R1 + R2 Example: two 100  light bulbs are connected (a) in series and (b) in parallel to a 24 V battery. What is the current through each bulb? For which circuit will the bulbs be brighter? (a) Series combination. R1 100 R2 100 Req = R1 + R2 + - V = I Req I V = 24 V V = I (R1 + R2) I = V / (R1 + R2) = 24 V / (100  + 100 ) = 0.12 A

(b) Parallel combination. V Example: two 100  light bulbs are connected (a) in series and (b) in parallel to a 24 V battery. What is the current through each bulb? For which circuit will the bulbs be brighter? I1 R1 (b) Parallel combination. V I2 R2 V + - I V = 24 V I (because R1 = R2)

Example: two 100  light bulbs are connected (a) in series and (b) in parallel to a 24 V battery. What is the current through each bulb? For which circuit will the bulbs be brighter? To answer the question, we must calculate the power dissipated in the bulbs for each circuit. The more power “consumed,” the brighter the bulb. In other words, we use power as a proxy for brightness.

(a) Series combination. R1 100 R2 100 Example: two 100  light bulbs are connected (a) in series and (b) in parallel to a 24 V battery. What is the current through each bulb? For which circuit will the bulbs be brighter? (a) Series combination. R1 100 R2 100 We know the resistance and current through each bulb, so for each bulb: + - I V = 24 V P = I2R P = (0.12 A)2 (100 ) P = 1.44 W

(b) Parallel combination. V Example: two 100  light bulbs are connected (a) in series and (b) in parallel to a 24 V battery. What is the current through each bulb? For which circuit will the bulbs be brighter? I1 R1 (b) Parallel combination. V We know the potential difference across each bulb, so for each bulb: I2 R2 V I I P = V2 / R + - P = (24 V)2 / ( 100 ) V = 24 V P = 5.76 W We also know each current, so we could have used P = I2R.

The bulbs in parallel are brighter. Example: two 100  light bulbs are connected (a) in series and (b) in parallel to a 24 V battery. What is the current through each bulb? For which circuit will the bulbs be brighter? I1 R1 R1 100 R2 100 V I2 R2 + - V I V = 24 V I I + - V = 24 V Compare: Pseries = 1.44 W Pparallel = 5.76 W The bulbs in parallel are brighter.

This is what you see if you connect 40 W bulbs directly to a 120 V outlet. (DO NOT TRY AT HOME.) Off On

Power = Energy Transformed / Time Energy Transformed = Power * Time Homework Hints (usefulness depends on what homework is assigned this semester) Power = Energy Transformed / Time Energy Transformed = Power * Time Do you remember this Phys. 1135 equation? Q (energy) = m c T

A Toy to Play With http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc