Answer: E. 79. What is the voltage across points A and B (Fig.6)?

 

79. What is the voltage across points A and B (Fig.6)?

A. 5V B. – 5V C. 29V D. 1V E. – 1V

 

Solution:

V_AB=2+15–12=5V

Answer: A.

Fig.6

 

 

 

80. What is the correct equation according to KCL for node B for the circuit below?

A. i₃+i₂+i_S2=0 B. i₃-i₂+i_S2=0 C. i₃-i₂-i_S2=0 D. - i₃+i₂+i_S2=0 E.- i₃+i₂-i_S2=0

Solution:

KCL states that the algebraic sum of the currents entering a node is zero at every instant. According to KCL equation for node B is following:

– i₃ + i₂ – i_S2 = 0

Answer: E.

 

 

81. When ideal switch is closed, the voltage is equal to______ and current is___, but when it is open current is equal to _____ and voltage is _________.

 

A. any value, 0, any value,0 B. any value, 0, 0, any value

C. any value, any value, 0, 0 D. 0, any value, 0, any value

E. 0, any value, any value, 0

 

Solution:

When switch is closed, it’s a short circuit: v=0 when R=0 and i=any value.

When switch is opened it’s an open circuit: i=0 when R=∞ and v=any value.

Answer: D.

 

 

82. What element(s) can be used as voltage divider?

A. resistor B. diode C. Transistor D. potentiometer E. resistor, potentiometer

Answer: E.

 

83. When star-connected circuit is balanced and transformed into the delta, the value of side of equivalent delta is equal to ___ of the star’s ray.

A.3 B.1/3 C.2/3 D.1/2 E. 2

 

Solution:

Let all resistors in star-connected circuit have the value R, then the value of side of equivalent delta is

, which is equal to 3 times of the star’s ray.