Answer: C.
13. What is the value of VX for the circuit in Fig.5? A. 10V B.5V C.-5V D.-10V E. –15V
Solution: Let’s consider the node at the center of the circuit. Current from left branch is I1=8/2*103= 4mA. By KCL: 4+5–10–ix=0. ix= –1mA. Vx=10*103*(–1*10-3)= –10V Answer: D.
Fig.5 14. What is the value of ix for the circuit in Fig.5? A. 1mA B.-1mA C.5mA D.-2mA E.2mA Solution: Let’s consider the node at the center of the circuit. Current from left branch is I1=8/2*103= 4mA. By KCL: 4+5–10–ix=0. ix= –1mA. Answer: B.
15. What i-v characteristic corresponds to resistor? Picture_____ A.A B.B C.C D.D E.E A B C D E
Solution: We can see i-v characteristic of resistor from Ohm’s Law: V=IR. Answer: C.
16. What statement is correct? A. A diode can be used as a voltage divider. B. Linear element of the circuit is such one whose characteristic is a straight line. C. Interface between a source and a load is one-port network device. D. Bilateral element of the circuit is such one whose i-v characteristic curve has odd symmetry about the origin. E. Bilateral element of the circuit is such one whose i-v characteristic curve has even symmetry about the origin.
Solution: A – wrong. Only resistor and potentiometer can be used as voltage dividers. B – wrong. Linear means that the defining characteristic is a straight line through the origin. Elements whose characteristics do not pass through the origin (even if it’s a straight line) or are not a straight line are said to be nonlinear. C – wrong. D – correct. E – wrong. Bilateral means that the i-v characteristic curve has odd symmetry about the origin.
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