Answer: D. A. The sum of currents at a node is 0

 

3. KCL states that

A. The sum of currents at a node is 0. B. the sum of currents in a loop is 0.

C. The algebraic sum of currents at a node is 0.

D. The algebraic sum of currents in a loop is 0.

E. The algebraic sum of currents at a node is 0 at every instant.

 

Solution:

Kirchhoff’s current law (KCL) is based on the principle of conservation of charge and states that the algebraic sum of currents at a node is 0 for every instant.

Answer: E.

 

4. The unit of current is

A. V B. J C. A D. F E. C

 

Solution:

Current is defined as dq/dt and is a measure of the flow of electrical charge. The unit of current is Ampere (A). 1 A = 1 Coulomb/second.

Answer: C.

 

5. We know R_A, R_B, and R_C for delta. What is R₂ for delta-to-star transformation? (see Fig. 1).

A. R₂ = (R_A + R_B + R_C)/ R_A B. R₂ = R_BR_C/ (R_A + R_B + R_C)

C. R₂ = R_AR_C/ (R_A + R_B + R_C) D. R₂ = R_BR_A/ (R_A + R_B + R_C)

E. R₂ = R_BR_C/ (R_A + R_B)

 

 

Fig.1

 

Solution:

Answer: C.

 

6. What is the value of current I? (see Fig.2)

A. 2A B. – 1A C. 5A D. 1A E. – 2A

 

Solution:

Both current sources are directed in the same direction, so the resulting current source will be presented as the sum I=2+3=5A.

Answer: C.

 

 

Fig. 2

7. Define V_A for the circuit in Fig.3. V_A=

A.V_S*R4/(R₁+R₄) B. V_S*R₂/(R₁+R₄) C. V_S*R₃/(R₁+R₄) D. V_S*R₃/(R₁+R₃) E. V_S*R₄/(R₁+R₃)

 

Solution:

By Voltage Division Rule:

Answer: D.

 

 

Fig.3

 

8. For the circuit in Fig.4 define V₀. Fig.4

A. 12V B.6V C.4V D.3V E. 10V

 

Solution:

Equivalent circuit will be

 

 

Now by Voltage Division Rule we have

V