1 of 11. Current Concept Tests.


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1 1 of 11 Current Concept Tests. CRKT1. ote 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! CRKT2. A 1Ω resistor is placed in parallel with a 10,000 Ω resistor as shown. 1 Ω I 10,000 Ω 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 1 Answer: a little less than 1Ω. You could use the formula R tot =, R1 R2 1 R = = Ω. Or, just think: 1Ω is a very low resistance compared to Ω, 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.
2 2 of 11 CRKT3. The circuit below consists of a battery attached to two resistors in series. Resistor R 1 is variable. When R 1 is decreased, the voltage 2 across R 2... A) increases B) decreases C) stays the same. 1 R 1 I 1 R 2 I 2 2 Answer: increases. This is easiest to see if you let R 1 go all the way to zero! Kirchhoff's 2nd Law says that the battery voltage = The voltage drop is split between resistors 1 and 2. So 2 is smaller than the battery voltage. But if R 1 =0, then the full battery voltage is across R 2. 2 increased as R 1 goes down. CRKT4. Two resistors R 1 and R 2 are hooked to a battery in parallel. R 1 is twice as large as R 2. How does the current I B from the battery compare to the current I 1 though R 1? (Hint: I B = I 1 + I 2.) I B R 1 I 1 = 2 R 2 R 2 I 2 A) I B = I 1 B) I B =2I 1 C) I B =3I 1 D) I B =4I 1 E) None of these. Answer: I B =3I 1 The voltage across R 1 is the same as the voltage across R 2, but R 1 is twice as large as R 2 so I 1 is half the size of I 2 (since I=/R, same, R twice as big means I half as large.) So if I 1 = 1A, I 2 would be 2A, and I tot =I B = I 1 + I 2 = 1A + 2A = 3A, which is three times as large as I B. I B =3I 1.
3 3 of 11 CRKT5. The four light bulbs shown are identical. Which circuit puts out more light? (Hint: more power = more light). = 12 = 12 (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 2 from the battery is P =. Smaller R tot (with fixed ) results in a larger power P. R tot So (A) puts out more light.
4 4 of 11 CRKT6. 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.) 2 A) Bulb 1 gets brighter B) Bulb 1 gets dimmer. C) Its brightness remains the same. 1 3 (Hint: What happens to the current from the battery when bulb 2 burns out.) = 12 Answer: When bulb 2 burns outs, the filament inside breaks and R 2 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 R tot, the current from the battery I tot = /R tot is reduced. Less current from the battery means less current through bulb 1, less light. Bulb 1 gets dimmer, thus B).
5 5 of 11 CRKT7. 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? 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. B C = 12 Answer: If each light bulb has the same resistance R, the series resistance of B and C is 2R. Power P= 2 /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. CRKT8. Two glowing light bulbs are in a batteryoperated 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. CRKT9. 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 120.) A) both have same brightness 40W 60W B) 40W is brighter C) 60W is brighter
6 6 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 120. So ( 120) 2 Plabeled =. So IF =120, THEN higher R means lower P. The bulb with the R smaller P labeled 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 = I 2 R. CRKT10. If you wanted to measure the current through the battery, where in the circuit would you place an ammeter? (A), (B), (C), (D), or (E) None of these will work. (A) R 2 R 1 (D) R 3 (C) (B) If you wanted to measure the current through resistor R 2, 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 this point. You wouldn t want to use D) because it is the set up for measuring the voltage across the battery, B) and C) have only part of the total current running through them. For the current running through R 2, use location (B). Only the current going through R 2 runs though this point. (A) is the current through R 1 and is the total current. (C) is the current through R 3 only.
7 7 of 11 CRKT11. A circuit with two batteries is shown below. The directions of the currents have been chosen (guessed) as shown. Which is the correct current equation for this circuit? A) I 2 = I 1 + I 3 B) I 1 = I 2 + I 3 C) I 3 = I 1 + I 2 E) None of these. I 1 R 1 1 R 2 2 I2 Loop 1 R 3 I 3 Answer: C), I 3 = I 1 + I 2. Consider the upper junction (marked with a dot). The current into the junction is I 3 (current arrow pointing towards the junction means current into the junction). The current out of the junction is I 1 +I 2. Kirchhoff's 2nd law says total current in equals total current out. You get exactly the same answer if you consider the lower junction. CRKT12. Which equation below is the correct equation for Loop 1? A) 2 + I 1 R 1 I 2 R 2 = 0 B) 2 + I 1 R 1 I 2 R 2 = 0 C) 2 + I 1 R 1 + I 2 R 2 = 0 D) 2 + I 1 R 1 + I 2 R 2 = 0 E) None of these. I 1 R 1 1 R 2 2 I2 Loop 1 R 3 I 3 Answer: As we move around loop 1, we go through the battery from the (+) to the ( )
8 8 of 11 side, a voltage drop so the voltage change is  2. Then we go through R 1 from the low side to the high side (since we are moving against the current flow I 1 ): a voltage rise so the change is +I 1 R 1. Finally, we go through R 2, in the same direction as the current I 2, so we have a voltage drop and a change of I 2 R 2. The total change as we moved around the loop must be zero, since we finished at the point where we started. 2 + I 1 R 1  I 2 R 2 = 0. CRKT13. What happens to the (identical) bulbs when the switch is closed? A) A glows, B changes B) A glows, B stays same C) A stays same, B changes D) A stays same, B stays same A switch 12 B 12 Answer: This is a tough one! A stays the same, B stays same. After the switch is closed, the voltage on both the right and left sides of bulb A is 12. Since there is no voltage difference across A, no current flows through A. The voltage across bulb B is FIXED by the 12 battery that it is in parallel with.
9 9 of 11 CRKT14. A flashlight requires 2 AA (1.5) batteries, and is arranged as shown. Notice that the switch is open. Which statement is true.? A) The bulb has 1.5 across it, and glows B) The bulb has 3 across it, and glows C) The bulb has 3 across it, and is dark D) The bulb has 0 across it, and is dark E) The bulb has 0 across it, and glows + AA AA Answer: The bulb has 0 across it, and is dark Which statement is true if the switched is closed? Answer: The bulb has 3 across it, and glows Which statement is true if the switched is closed and one of the two batteries is reversed? Answer: The bulb has 0 across it, and is dark. If one battery is reversed, the batteries are fighting each other. CRKT15. A capacitor in an RC circuit is initially charged up to a voltage of 10 and is then discharged through an R=10Ω resistor as shown. The switch is closed at time t= F I 10Ω What is the current through the resistor, immediately after the switch is closed, at time t = 0.0+ s? A) 1A B) 0.5A C) 1/e A = 0.37A D) None of these. What is the time constant for this circuit? A) 0.01 s B) 0.1 s C) 1 s D) 10 s D) None of these. What is the current through the resistor at time t = 0.2 s? A) 1A B) 0.5A C) 1/e A = 0.37A D) None of these. Answers: Right after the switch is closed, the current thru R is I = /R = 10/10ohm = 1A. The time constant for this circuit is RC=10Ω 0.010F = 0.10 sec. So at time t=0.2
10 10 of 11 sec, two time constants have passed. After one time constant, the voltage, charge, and current have all decreased by a factor of e. After two time constants, everything has fallen by e 2. The initial current is 1A. So after two time constants, the current is 1/e 2 A = 0.135A. So the answer is none of these. CRKT16. An RC circuit is shown below. Initially the switch is open and the capacitor has no charge. At time t=0, the switch is closed. What is the voltage across the capacitor immediately after the switch is closed (time = 0+)? R = 10Ω = 10 C = F R = 10Ω A) Zero B) 10 C) 5 D) None of these. What is the voltage across the resistor on the far right (in parallel with the capacitor) at time = 0+? A) Zero B) 10 C) 5 D) None of these. What is the initial current "through" the capacitor (immediately after the switch is closed)?a) 1A B) zero C) 0.5A D) None of these. A long time after the switch has been closed, what is the voltage across the capacitor? A) 5 B) 10 C) zero D) None of these. Answers: Right after the switch is closed, C = 0. Before the switch is closed, the charge Q on the capacitor is zero and the voltage across the capacitor = = Q/C = 0. Right after the switch is closed, the charge has not had time to build up on the capacitor and the charge and voltage are still zero. Right after the switch is closed, R = 0. The capacitor and the resistor on the right are in parallel and therefore have the same voltage drop across them. Since the capacitor is acting like a wire there is no voltage drop across it, which is the same for the resistor on the right. The initial current thru the capacitor is 1A. Initially, when the capacitor has zero charge, it behaves like a shortcircuit (zero resistance) because it is easy to put charge on an uncharged capacitor. The circuit is then, effectively...
11 11 of 11 R = 10Ω =10 I=/R=10/10=1A The capacitor acting like a zero resistance wire which all the current flows through (initially). The other resistor is not involved in the (initial) current flow. After a long time, the voltage across the capacitor is 5. After a long time, the capacitor becomes fully charged, and current stops flowing through it. When this happens it behaves like an infinite resistor, and the circuit is effectively.. R = 10Ω =10 R = 10Ω C= F I=/2R=10/20=0.5A As far as current flow is concerned, the capacitor is gone. The voltage across the capacitor is the same as the voltage on the right, since they are in parallel. The voltage across each resistor is 5. Same R's, same I in each, so the voltage across each must be the same and they must add up to 10.
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