1 of 11 Current Concept Tests. CRKT-1. Vote TRUE(A) if both statements below are always true. Otherwise, vote FALSE(B). • For resistors in series, the current through each resistor is the same. • For resistors in parallel, the voltage across each resistor is the same. Answer: True! CRKT-2. A 1Ω resistor is placed in parallel with a 10,000 Ω resistor as shown. The total, equivalent resistance of these two resistor in parallel is closest to... A) a little less than 1Ω B) a little more than 1Ω. C) 5000 Ω D) a little less than 10000Ω E) a little more than 10000Ω Answer: a little less than 1Ω. You could use the formula RRRtot=+11112, R =+=1110000110 9990. Ω. Or, just think: 1Ω is a very low resistance compared to 10000Ω, so almost all the current will flow through the 1Ω resistor, the circuit will behave almost as if the 10000Ω resistor is not present, and the equivalent resistance is close to 1Ω. The question is: is the equivalent resistance a little less or a little greater than 1Ω. Adding the 10000Ω resistor in parallel provides another current path of the flow of charge. More flow means lower resistance. 1 Ω I 10,000 Ω M.Dubson, Phys1120 Notes ©University of Colorado at Boulder2 of 11 CRKT-3. The circuit below consists of a battery attached to two resistors in series. Resistor R1 is variable. When R1 is decreased, the voltage V2 across R2 ... A) increases B) decreases C) stays the same. Answer: increases. This is easiest to see if you let R1 go all the way to zero! Kirchhoff's 2nd Law says that the battery voltage V=V1+V2. The voltage drop is split between resistors 1 and 2. So V2 is smaller than the battery voltage V. But if R1=0, then the full battery voltage V is across R2. V2 increased as R1 goes down. CRKT-4. Two resistors R1 and R2 are hooked to a battery in parallel. R1 is twice as large as R2. How does the current IB from the battery compare to the current I1 though R1? (Hint: IB = I1+ I2.) A) IB = I1 B) IB=2I1 C) IB=3I1 D) IB=4I1 E) None of these. Answer: IB=3I1 The voltage across R1 is the same as the voltage across R2, but R1 is twice as large as R2 so I1 is half the size of I2 (since I=V/R, same V, R twice as big means I half as large.) So if I1 = 1A, I2 would be 2A, and Itot=IB = I1 + I2 = 1A + 2A = 3A, which is three times as large as IB. IB=3I1. R1 I1V R2 I2= 2 R2 IB R2 R1 V1 I1 V2 I2V M.Dubson, Phys1120 Notes ©University of Colorado at Boulder3 of 11 CRKT-5. The four light bulbs shown are identical. Which circuit puts out more light? (Hint: more power = more light). (A) (B) (C) They both put out the same amount of light. Answer: The total equivalent resistance which the battery in the (A) circuit sees is R/2 (two resistors, each of resistance R in parallel) The total equivalent resistance which the battery in the (B) circuit sees is 2R (two resistors in series). The total power coming from the battery is PVRtot=2. Smaller Rtot (with fixed V) results in a larger power P. So (A) puts out more light. V = 12V V = 12V M.Dubson, Phys1120 Notes ©University of Colorado at Boulder4 of 11 CRKT-6. In the circuit below, what happens to the brightness of bulb 1, when bulb 2 burns out? (When a bulb burns out, its resistance becomes infinite.) A) Bulb 1 gets brighter V = 12V 2 B) Bulb 1 gets dimmer. C) Its brightness remains the same. 3 1 (Hint: What happens to the current from the battery when bulb 2 burns out.) Answer: When bulb 2 burns outs, the filament inside breaks and R2 becomes infinitely large. The total equivalent resistance which the battery sees increases (since bulb 2 is gone, there are fewer paths for the current flow, so less flow, more total resistance.) Since the battery sees a larger Rtot, the current from the battery Itot = V/Rtot is reduced. Less current from the battery means less current through bulb 1, less light. Bulb 1 gets dimmer, thus B). M.Dubson, Phys1120 Notes ©University of Colorado at Boulder5 of 11 CRKT-7. The three light bulbs A, B, and C are identical. How does the brightness of bulbs B and C together compare with the brightness of bulb A? V = 12V A A) Total power in B+C = power in A. B) Total power in B+C > power in A. C) Total power in B+C < power in A. C B Answer: If each light bulb has the same resistance R, the series resistance of B and C is 2R. Power P=V2/R. Larger total resistance for the B/C pair means less power. Total power in B+C < power in A, therefore the correct answer is C. CRKT-8. Two glowing light bulbs are in a battery-operated circuit. Light bulb A has greater resistance than light bulb B. Which light bulb is brighter? A B C) Depends on the circuit. Answer: Depends on the circuit. If the resistors (the light bulbs) are in series, the larger R will be brighter. But if the resistors are in parallel, the smaller R will be brighter. CRKT-9. Two light bulbs are in series attached to a battery as shown. The bulbs are marked 40W and 60W. Which bulb is brighter? (Hints: More power = brighter. When light bulbs are in series, they have the same current. Light bulbs are intended to operate at 120V.) A) both have same brightness B) 40W is brighter C) 60W is brighter V 40W 60W M.Dubson, Phys1120 Notes ©University of Colorado at Boulder6 of 11 Answer: This is a tricky one! The answer is that the “40W” bulb is brighter. The bulbs are labeled assuming that they would be used with a voltage of 120V. So ()2labeled120VPR=. So IF V=120V, THEN higher R means lower P. The bulb with the smaller Plabeled has the larger R. The “40W” bulb has higher R than the “60W” bulb. These two light bulbs are in series (and light bulbs are not intended to be used this way). In series, the current is the same, so the larger R produces the larger P = I2 R. CRKT-10. If you wanted to measure the current through the battery, where in the circuit would you place an ammeter? V R3 R2 (B) (A) (A), (B), (C), (D), or (E) None of these will work. R1 (D) (C) If you wanted to measure the current through resistor R2, where would you place and ammeter? (A), (B), (C), (D), or (E) None of these will work. Answers: Use location (A) to measure battery current, since all of the current that is going through the battery is also running through
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