# Clicker Session Currents, DC Circuits

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1 Clicker Session Currents, DC Circuits

2 Wires A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance? 1) it decreases by a factor 4 2) it decreases by a factor 2 3) it stays the same 4) it increases by a factor 2 5) it increases by a factor 4

3 Wires A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance? 1) it decreases by a factor 4 2) it decreases by a factor 2 3) it stays the same 4) it increases by a factor 2 5) it increases by a factor 4 Keeping the volume (= area x length) constant means that if the length is doubled, the area is halved. L Since R = ρ, this increases the resistance by four. A

4 Series Resistors I Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? 1) 12 V 2) zero 3) 3 V 4) 4 V 5) you need to know the actual value of R 9 V

5 Series Resistors I Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? Since the resistors are all equal, the voltage will drop evenly across the 3 resistors, with 1/3 of 9 V across each one. So we get a 3 V drop across each. 1) 12 V 2) zero 3) 3 V 4) 4 V 5) you need to know the actual value of R 9 V Follow-up: What would be the potential difference if R= 1 Ω, 2 Ω, 3 Ω

6 Series Resistors II 1) 12 V In the circuit below, what is the voltage across R 1? 2) zero 3) 6 V 4) 8 V 5) 4 V R 1 = 4 Ω R 2 = 2 Ω 12 V

7 Series Resistors II 1) 12 V In the circuit below, what is the voltage across R 1? 2) zero 3) 6 V 4) 8 V 5) 4 V The voltage drop across R 1 has to be twice as big as the drop across R 2. This means that V 1 = 8 V and V 2 = 4 V. Or else you could find the current I = V/R = (12 V)/(6 Ω) = 2 A, then use Ohm s Law to get voltages. R 1 = 4 Ω R 2 = 2 Ω 12 V

8 Parallel Resistors I 1) 10 A In the circuit below, what is the current through R 1? 2) zero 3) 5 A 4) 2 A 5) 7 A R 2 = 2 Ω R 1 = 5 Ω 10 V

9 Parallel Resistors I 1) 10 A In the circuit below, what is the current through R 1? 2) zero 3) 5 A 4) 2 A 5) 7 A The voltage is the same (10 V) across each resistor because they are in parallel. Thus, we can use Ohm s Law, V 1 = I 1 R 1 to find the current I 1 = 2 A. R 2 = 2 Ω R 1 = 5 Ω Follow-up: What is the total current through the battery? 10 V

10 Parallel Resistors II Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit? 1) increases 2) remains the same 3) decreases 4) drops to zero

11 Parallel Resistors II Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit? 1) increases 2) remains the same 3) decreases 4) drops to zero As we add parallel resistors, the overall resistance of the circuit drops. Since V = IR, and V is held constant by the battery, when resistance decreases, the current must increase. Follow-up: What happens to the current through each resistor?

12 Which of these diagrams represent the same circuit? Diagrams A. a and b B. a and c C. b and c D. a, b, and c E. a, b, and d

13 Which of these diagrams represent the same circuit? Diagrams A. a and b B. a and c C. b and c D. a, b, and c E. a, b, and d The three elements are in parallel their ends are connected by conducting wires. The order of the elements and length of the connecting wires are immaterial.

14 Circuit I Three lightbulbs, A, B and C are in the circuit shown. When the switch is closed, lightbulb A will: 1) glow brighter than before 2) glow just the same as before 3) glow dimmer than before 4) go out completely 5) explode

15 Circuit I Three lightbulbs, A, B and C are in the circuit shown. When the switch is closed, lightbulb A will: 1) glow brighter than before 2) glow just the same as before 3) glow dimmer than before 4) go out completely 5) explode Since the total resistance through bulbs B and C is now less than the resistance through B alone, the total resistance of the circuit decreases. This means that the current through bulb A increases. Follow-up: What happens to bulb B?

16 Circuits II The lightbulbs in the circuit below are identical with the same resistance R. Which circuit produces more light? (brightness power) 1) circuit 1 2) circuit 2 3) both the same 4) it depends on R

17 Circuits II The lightbulbs in the circuit below are identical with the same resistance R. Which circuit produces more light? (brightness power) 1) circuit 1 2) circuit 2 3) both the same 4) it depends on R In #1, the bulbs are in parallel, lowering the total resistance of the circuit. Thus, circuit #1 will draw a higher current, which leads to more light, because P = I V.

18 More Circuits I What happens to the voltage across the resistor R 1 when the switch is closed? The voltage will: 1) increase 2) decrease 3) stay the same R 1 V S R 2 R 3

19 More Circuits I What happens to the voltage across the resistor R 1 when the switch is closed? The voltage will: 1) increase 2) decrease 3) stay the same With the switch closed, the addition of R 1 R 2 to R 3 decreases the equivalent resistance, so the current from the battery increases. This will cause an increase in the voltage across R 1. V S R 2 R 3 Follow-up: What happens to the current through R 3?

20 More Circuits II What happens to the voltage across the resistor R 4 when the switch is closed? 1) increases 2) decreases 3) stays the same R 1 V S R 2 R 3 R 4

21 More Circuits II What happens to the voltage across the resistor R 4 when the switch is closed? 1) increases 2) decreases 3) stays the same We just saw that closing the switch causes an increase in the voltage across R 1 (which is V AB ). The A R 1 S B voltage of the battery is constant, V R 3 R 4 so if V AB increases, then V BC must R 2 decrease! C Follow-up: What happens to the current through R 4?

22 Even More Circuits Which resistor has the greatest current going through it? Assume that all the resistors are equal. 1) R 1 2) both R 1 and R 2 equally 3) R 3 and R 4 4) R 5 5) all the same V

23 Even More Circuits Which resistor has the greatest current going through it? Assume that all the resistors are equal. 1) R 1 2) both R 1 and R 2 equally 3) R 3 and R 4 4) R 5 5) all the same The same current must flow through left and right combinations of resistors. On the LEFT, the current splits equally, so I 1 = I 2. On the RIGHT, more current will V go through R 5 than R 3 + R 4 since the branch containing R 5 has less resistance. Follow-up: Which one has the smallest voltage drop?

24 Dimmer When you rotate the knob of a light dimmer, what is being changed in the electric circuit? 1) the power 2) the current 3) the voltage 4) both (1) and (2) 5) both (2) and (3)

25 Dimmer When you rotate the knob of a light dimmer, what is being changed in the electric circuit? 1) the power 2) the current 3) the voltage 4) both (1) and (2) 5) both (2) and (3) The voltage is provided at 120 V from the outside. The light dimmer increases the resistance and therefore decreases the current that flows through the lightbulb. Follow-up: Why does the voltage not change?

26 Space Heaters Two space heaters in your living room are operated at 120 V. Heater 1 has twice the resistance of heater 2. Which one will give off more heat? 1) heater 1 2) heater 2 3) both equally

27 Space Heaters Two space heaters in your living room are operated at 120 V. Heater 1 has twice the resistance of heater 2. Which one will give off more heat? 1) heater 1 2) heater 2 3) both equally Using P = V 2 / R, the heater with the smaller resistance will have the larger power output. Thus, heater 2 will give off more heat. Follow-up: Which one carries the greater current?

28 Junction Rule What is the current in branch P? 1) 2 A 2) 3 A 3) 5 A 4) 6 A 5) 10 A 5 A 8 A P 2 A

29 Junction Rule 1) 2 A What is the current in branch P? 2) 3 A 3) 5 A 4) 6 A 5) 10 A The current entering the junction in red is 8 A, so the current leaving must also be 8 A. One exiting branch has 2 A, so the other branch (at P) must have 6 A. 5 A S 8 A junction 2 A P 6 A

30 Kirchhoff s Rules 1) 2 I 1 2I 2 = 0 Which of the equations is valid 2) 2 2I 1 2I 2 4I 3 = 0 for the circuit below? 3) 2 I 1 4 2I 2 = 0 4) I 3 4 2I = 0 5) 2 I 1 3I 3 6= 0 1 Ω I 2 2 Ω 2 V 6 V 4 V I I Ω 3 Ω

31 Kirchhoff s Rules 1) 2 I 1 2I 2 = 0 Which of the equations is valid 2) 2 2I 1 2I 2 4I 3 = 0 for the circuit below? 3) 2 I 1 4 2I 2 = 0 4) I 3 4 2I = 0 5) 2 I 1 3I 3 6= 0 Eqn. 3 is valid for the left loop: The left battery gives +2V, then there is a drop through a 1Ω resistor with current I 1 flowing. Then we go through the middle battery (but from + to!), which gives 4V. Finally, there is a drop through a 2Ω resistor with current I 2. I 2 1 Ω 2 Ω 2 V 6 V 4 V I I Ω 3 Ω

32 The time constant for the discharge of this capacitor is RC Circuits A. 1 s. B. 2 s. C. 4 s. D. 5 s. E. The capacitor doesn t discharge because the resistors cancel each other.

33 The time constant for the discharge of this capacitor is RC Circuits A. 1 s. B. 2 s. C. 4 s. D. 5 s. E. The capacitor doesn t discharge because the resistors cancel each other.

34 Circuits III The three lightbulbs in the circuit all have the same resistance of 1 Ω. By how much is the brightness of bulb B greater or smaller than the brightness of bulb A? (brightness power) 1) twice as much 2) the same 3) 1/2 as much 4) 1/4 as much 5) 4 times as much A B C 10 V

35 Circuits III The three lightbulbs in the circuit all have the same resistance of 1 Ω. By how much is the brightness of bulb B greater or smaller than the brightness of bulb A? (brightness power) 1) twice as much 2) the same 3) 1/2 as much 4) 1/4 as much 5) 4 times as much A We can use P = V 2 /R to compare the power: P A = (V A ) 2 /R A = (10 V) 2 /1 Ω = 100 W B C P B = (V B ) 2 /R B = (5 V) 2 /1 Ω = 25 W 10 V Follow-up: What is the total current in the circuit?

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