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Power Transfer For the circuit shown, plot of PL, the power dissipated in the load Resistance RL, is given as a function of RL. 10 Q2- Rs P (Q15); The value of resistance Rs (in (2) is: b)50 a) 30 c)80 d) 40

Power Transfer For the circuit shown, plot of PL, the power dissipated in the load Resistance RL, is given as a function of RL. 10 Q2- Rs P (Q15); The value of resistance Rs (in (2) is: b)50 a) 30 c)80 d) 40

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Q 15 please
Wheatstone Bridge: (910); In the circuit shown. What is the equivalent resistance of the circuit? a) 20 b) 60 c) 40 d) 120 Kirchhoff's Laws The ammeter shown in the circuit reads 1.6 A (Q13): Calculate the currents 11 and 12 as (11,12) ? a) (0.3,1.3)A b) (1,0.6)A c) (0.6,1)A Galvanometer (011); In the shown figure, a galvanometer with (full scale 5 mA, R=1 KO). The value of Rx (in k0) that makes the galvanometer reads a voltage of order of 30 V is: 460 (Q14); Calculate e? a) 9.2V 10 2 Rs b) 8.4V c) 10V E a) 2 b) 10 c) 15 d) 12 Rx (Q12): To convert the Galvanometer into an ammeter we must connect a a)parallel small resistor b) parallel large resistor c) series small resistor d) parallel large resistor 100 d) 12V Power Transfer For the circuit shown, plot of PL, the power dissipated in the load Resistance RL, is given as a function of RL. 50 (Q15); The value of resistance Rs (in ) is: b)50 c)80 a) 30 d) 40 (Q16): Calculate the emf of the battery E (in Volt)? a) 8.34 c) 3.09 d) 2.76 b) 6.01 20 d) (1.3,0.3)A Allt 25V 40 7.000 W PL (mW) 10 201 20 3.000 W W 2.000 15.0V
Transcribed Image Text:Wheatstone Bridge: (910); In the circuit shown. What is the equivalent resistance of the circuit? a) 20 b) 60 c) 40 d) 120 Kirchhoff's Laws The ammeter shown in the circuit reads 1.6 A (Q13): Calculate the currents 11 and 12 as (11,12) ? a) (0.3,1.3)A b) (1,0.6)A c) (0.6,1)A Galvanometer (011); In the shown figure, a galvanometer with (full scale 5 mA, R=1 KO). The value of Rx (in k0) that makes the galvanometer reads a voltage of order of 30 V is: 460 (Q14); Calculate e? a) 9.2V 10 2 Rs b) 8.4V c) 10V E a) 2 b) 10 c) 15 d) 12 Rx (Q12): To convert the Galvanometer into an ammeter we must connect a a)parallel small resistor b) parallel large resistor c) series small resistor d) parallel large resistor 100 d) 12V Power Transfer For the circuit shown, plot of PL, the power dissipated in the load Resistance RL, is given as a function of RL. 50 (Q15); The value of resistance Rs (in ) is: b)50 c)80 a) 30 d) 40 (Q16): Calculate the emf of the battery E (in Volt)? a) 8.34 c) 3.09 d) 2.76 b) 6.01 20 d) (1.3,0.3)A Allt 25V 40 7.000 W PL (mW) 10 201 20 3.000 W W 2.000 15.0V
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