College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Two resistors connected in series have an equivalent resistance of 644 Ω. When they are connected in parallel, their equivalent resistance is 118 Ω. Find the resistance of each resistor.
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- A battery has a terminal voltage of 13.5 V when no current flows. Its internal resistance is 2.00 Ω. A 3.30-Ω resistor is connected across the battery terminals. What is the current through the 3.30-Ω resistor?arrow_forwardThree 7.50 Ω resistors are connected in series to a 22.0 V battery. What is the equivalent resistance (in Ω) of the circuit? What is the current (in A) in each resistor? (b) Three other 7.50 Ω resistors are all connected in parallel across a second 22.0 V battery. What is the equivalent resistance (in Ω) of this circuit? What is the current (in A) in each resistor in this circuit?arrow_forwardWhen resistors 1 and 2 are connected in series, the equivalent resistance is 17.1 Q. When they are connected in parallel, the equivalent resistance is 2.93 Q. What are (a) the smaller resistance and (b) the larger resistance of these two resistors? (a) Number i Units (b) Number i Unitsarrow_forward
- Three resistors having resistances of R1 = 1.76 Ω , R2 = 2.64 Ω and R3 = 4.84 Ω respectively, are connected in series to a 27.7 V battery that has negligible internal resistance. Find the total current through the battery, the voltage across each resistor, and the power dissipated in each resistor.arrow_forwardTwo resistors connected in series have an equivalent resistance of 620 Ω. When they are connected in parallel, their equivalent resistance is 122 Ω. Find the resistance of each resistor.arrow_forwardThree resistors having resistances of R1 = 1.76 Ω , R2 = 2.64 Ω and R3 = 4.84 Ω respectively, are connected in series to a 27.7 V battery that has negligible internal resistance. Find the equivalent resistance of the combination and the current in each resistor.arrow_forward
- Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?arrow_forwardThe circuit below contains two real batteries, each having significant internal resistance. Battery #1 has an EMF of 6.00 V and an internal resistance of 1.00 ohm. Battery #2 has an EMF of 12.00 V and an internal resistance of 2.00 ohms. These two batteries are connected in series with two resistors R, = 6.00 ohms and R,= 3.00 ohms. ri d (6.00,V) (1.00 0) R1 (6.00 N) R2 (3.00 N) a r2 E2 (2.00 Q) (12.00 V) 18. The current in the 12.00 Volt battery is equal to C. 0.500 A. A. 0.750 A. B. 1.00 A. D. 2.00 A. E. 1.50 A. 19. The potential difference across the terminals of battery #1 from point C to point d is equal to C. 7.00 V. A. 5.50 V. B. 6.00 V. D. 5.00 V. E. 6.50 V. 20. The potential difference across the terminals of battery #2 from point a to point b is equal to C. 13.00 V. A. 14.00 V. В. 10.00 V. D. 11.0 V. E. 12.00 V.arrow_forward(a)Calculate the equivalent resistance (in Ω) between points a and b. Ω (b)If a voltage of 41.0 V is applied between points a and b, find the current (in A) in each resistor. 12 Ω A 6.0 Ω A 5.0 Ω A 4.0 Ω A 8.0 Ωarrow_forward
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