On the left the current splits equally so i1 i2 on the

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Unformatted text preview: than R3 + R4 since the branch containing R5 has less resistance. R1 V R2 R4 R3 R5 ConcepTest 4.7 Junction Rule ConcepTest Junction a) 2 A What is the current in branch P? b) 3 A c) 5 A d) 6 A e) 10 A 5A P 8A 2A ConcepTest 4.7 Junction Rule Junction a) 2 A b) 3 A What is the current in branch P? c) 5 A d) 6 A e) 10 A The current entering the junction in red is 8 A, so the current red leaving must also be 8 A. One exiting branch has 2 A, so the exiting other branch (at P) must have 6 A. S 5A P 8A junction 2A 6A ConcepTest 4.9 Wheatstone Bridge ConcepTest An ammeter A is connected An a) I a) between points a and b in the b) I/2 b) I/2 circuit below, in which the four c) I/3 c) I/3 resistors are identical. The current identical The d) I/4 d) I/4 through the ammeter is: through e) zero a b V I ConcepTest 4.9 Wheatstone Bridge ConcepTest Wheatstone An ammeter A is connected An a) I a) between points a and b in the b) I/2 b) I/2 circuit below, in which the four c) I/3 c) I/3 resistors are identical. The current identical The d) I/4 d) I/4 through the ammeter is: through e) zero Since all resistors are identical, resistors a the voltage drops are the same voltage across the upper branch and the lower branch. Thus, the potentials at points a and b are potentials b also the same. Therefore, no same current flows. V I ConcepTest 4.10 ConcepTest More Kirchhoff’s Rules Kirchhoff’s a) 2 – I1 – 2I2 = 0 Which of the equations is valid Which b) 2 – 2I1 – 2I2 – 4I3 = 0 b) 2I 4I for the circuit below? c ) 2 – I 1 – 4 – 2 I2 = 0 c) d ) I 3 – 4 – 2 I2 + 6 = 0 d) e) 2 – I1 – 3I3 – 6 = 0 e) 1Ω I2 2Ω 6V 2V 2V 4V I1 1Ω I3 3Ω ConcepTest 4.10 More Kirchhoff’s Rules Kirchhoff’s a) 2 – I1 – 2I2 = 0 Which of the equations is valid b) 2 – 2I1 – 2I2 – 4I3 = 0 b) 2I 4I for the circuit below? c ) 2 – I 1 – 4 – 2 I2 = 0 c) d ) I 3 – 4 – 2 I2 + 6 = 0 d) e) 2 – I1 – 3I3 – 6 = 0 e) Eqn. 3 is valid for the left loop: Eqn. The left battery gives +2V, then there is a drop through a 1Ω resistor with current I1 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 I2. 1Ω I2 2Ω 6V 2V 2V 4V I1 1Ω I3 3Ω...
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