
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Question
a) Find the magnitude of the current through the 20.0 Ω resistor.
b) Find the direction of the current through the 15.0 Ω resistor. Circle one. Explain your answer below.
The current goes toward the lower right of the circuit OR toward the upper left of the circuit.
c) Find the voltage drop across the 10.0 Ω resistor.

Transcribed Image Text:20.0 2
12.0 V
15.0 2
5.00 2
7.00 V
3.00 V
10.0 2
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- A.) What is the potential difference between a and b? B.) What is the equivalent resistance of the circuit?arrow_forward2. The capacitor in the figure below is initially uncharged and the switch, in position c, is not connected to either side of the circuit. The switch is now flipped to position a for 10 ms, then to position b for 10 ms, and then brought back to position c. What is the final potential difference across the capacitor? 12 V 150 Ω ww a b 40 μF 150 Ωarrow_forwardAn electric circuit consisting of wires, batteries and resistors is shown. All currents have been labeled and assigned directions. a.) Which statement is a correct expression of Kirchhoff's Junction Rule for currents with labels and directions assigned on the above figure? I1 + I2 + I3 = 0 I1 - I2 + I3 = 0 None of these statements are correct. I1 - I2 - I3 = 0 I1 + I2 - I3 = 0 b.) Which statement is a correct expression of Kirchhoff's Loop Rule for the sum of voltage changes for the large outer loop? 18.0 V - (80.0 Ω)I1 + 12.0 V + (50.0 Ω)I3 + (80.0 Ω)I3 = 0 18.0 V - (80.0 Ω)I1 - 12.0 V + (50.0 Ω)I3 + (80.0 Ω)I3 = 0 None of these statements are correct. 18.0 V - (80.0 Ω)I1 + 12.0 V - (50.0 Ω)I3 - (80.0 Ω)I3 = 0 -18.0 V - (80.0 Ω)I1 - 12.0 V + (50.0 Ω)I3 + (80.0 Ω)I3 = 0 18.0 V - (80.0 Ω)I1 - 12.0 V - (50.0 Ω)I3 - (80.0 Ω)I3 = 0arrow_forward
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