# 8 - karna(pk4534 – HW 06 – li –(59050 1 This...

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Unformatted text preview: karna (pk4534) – HW 06 – li – (59050) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 3 Ω 4 Ω 6 Ω 9 Ω 11 Ω 2 Ω 1 Ω 2 Ω 11 V 23 V 37 V Find the magnitude of the current in the 11 V cell. Correct answer: 1 . 03145 A. Explanation: i 1 R 1 i 3 R 2 i 1 R 3 i 3 R 4 i 2 R 5 r 1 r 2 r 3 E 1 E 2 E 3 Let : E 1 = 11 V , E 2 = 23 V , E 3 = 37 V , R 1 = 3 Ω , R 2 = 4 Ω , R 3 = 6 Ω , R 4 = 9 Ω , R 5 = 11 Ω , r 1 = 2 Ω , r 2 = 1 Ω , and r 3 = 2 Ω . Basic Concepts: Kirchhoff’s Laws: summationdisplay V = 0 around a closed loop . summationdisplay I = 0 at a circuit junction . Solution: Applying Kirchhoff’s law to the outside loop and the lower loop we get 3 equa- tions in 3 unknowns; i.e. , E 1 − E 3 = ( R 3 + r 1 + R 1 ) i 1 + ( R 2 + r 3 + R 4 ) i 3 (1) E 2 − E 3 = ( R 5 + r 2 ) i 2 + ( R 2 + r 3 + R 4 ) i 3 (2) 0 = − i 1 − i 2 + i 3 . (3) Subtracting the first two equations, E 1 − E 2 = ( R 3 + r 1 + R 1 ) i 1 − ( R 5 + R 2 ) i 2 (4) Eliminating i 3 in equations (2) and (3), E 2 − E 3 = ( R 2 + r 3 + R 4 ) i 1 + ( R 2 + r 3 + R 4 + R 5 + r 2 ) i 2 (5) Multiply equation (4) by ( R 2 + r 3 + R 4 + R 5 + r 2 ) and equation (5) by ( R 5 + r 2 ): ( E 1 − E 2 ) ( R 2 + r 3 + R 4 + R 5 + r 2 ) = ( R 3 + r 1 + R 1 ) × ( R 2 + r 3 + R 4 + R 5 + r 2 ) i 1 − ( R 5 + r 2 ) × ( R 2 + r 3 + R 4 + R 5 + r 2 ) i 2 ( E 2 − E 3 ) ( R 5 + r 2 ) = ( R 2 + r 3 + R 4 )( R 5 + r 2 ) i 1 − ( R 5 + r 2 ) × ( R 2 + r 3 + R 4 + R 5 + r 2 ) i 2 Adding, E 1 ( R 2 + r 3 + R 4 + R 5 + r 2 ) − E 2 ( R 2 + r 3 + R 4 ) − E 3 ( R 5 + r 2 ) = [( R 3 + r 1 + R 1 )( R 2 + r 3 + R 4 + R 5 + r 2 ) + ( R 2 + r 3 + R 4 )( R 5 + r 2 )] i 1 karna (pk4534) – HW 06 – li – (59050) 2 Since a = R 2 + r 3 + R 4 + R 5 + r 2 = 4 Ω + 2 Ω + 9 Ω + 11 Ω + r 2 = 27 Ω , b = R 2 + r 3 + R 4 = 4 Ω + 2 Ω + 9 Ω = 15 Ω , c = R 5 + r 2 = 11 Ω + 1 Ω = 12 Ω , and d = ( R 3 + r 1 + R 1 ) a + b c = (6 Ω + 2 Ω + 3 Ω) (27 Ω) + (15 Ω) (12 Ω) = 477 Ω , we have E 1 a − E 2 b − E c = d i 1 i 1 = E 1 a − E 2 b − E c d = (11 V)(27 Ω) − (23 V)(15 Ω) 477 Ω − (37 V)(12 Ω) 477 Ω = − 1 . 03145 A , which has a magnitude of 1 . 03145 A . 002 10.0 points In the figure below the battery has an emf of 25 V and an internal resistance of 1 Ω . Assume there is a steady current flowing in the circuit. 3 μ F 8 Ω 6 Ω 1 Ω 25 V Find the charge on the 3 μ F capacitor. Correct answer: 30 μ C. Explanation: Let : R 1 = 8 Ω , R 2 = 6 Ω , r in = 1 Ω , V = 25 V , and C = 3 μ F . The equivalent resistance of the three resistors in series is R eq = R 1 + R 2 + r in = (8 Ω) + (6 Ω) + (1 Ω) = 15 Ω , so the current in the circuit is I = V R eq , and the voltage across R 2 is V 2 = I R 2 = R 2 R eq V = (6 Ω) (15 Ω) (25 V) = 10 V ....
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8 - karna(pk4534 – HW 06 – li –(59050 1 This...

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