Review Questions

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Review Questions

3.1 At node 1 in the circuit of Fig. 3.46, applying KCL gives:

(a)

$$2 + \frac{12 - v_1}{3} = \frac{v_1}{6} + \frac{v_1 - v_2}{4}$$

\n(b) $2 + \frac{v_1 - 12}{3} = \frac{v_1}{6} + \frac{v_2 - v_1}{4}$
\n(c) $2 + \frac{12 - v_1}{3} = \frac{0 - v_1}{6} + \frac{v_1 - v_2}{4}$
\n(d) $2 + \frac{v_1 - 12}{3} = \frac{0 - v_1}{6} + \frac{v_2 - v_1}{4}$

Figure 3.46

• For Review Questions 3.1 and 3.2.
• 3.2 In the circuit of Fig. 3.46, applying KCL at node 2 gives:

3.3 For the circuit in Fig. 3.47, v1 and v2 are related as:

(a)

$$v_1 = 6i + 8 + v_2$$

\n(b) $v_1 = 6i - 8 + v_2$
\n(c) $v_1 = -6i + 8 + v_2$
\n(d) $v_1 = -6i - 8 + v_2$

Figure 3.47

For Review Questions 3.3 and 3.4.

3.4 In the circuit of Fig. 3.47, the voltage v2 is:

(a) −8 V	(b) −1.6 V
(c) 1.6 V	(d) 8 V

3.5 The current i in the circuit of Fig. 3.48 is:

(a) −2.667 A	(b) −0.667 A
(c) 0.667 A	(d) 2.667 A

Figure 3.48

For Review Questions 3.5 and 3.6.

• 3.6 The loop equation for the circuit in Fig. 3.48 is:

  • (a) −10 + 4i + 6 + 2i = 0 (b) 10 + 4i + 6 + 2i = 0 (c) 10 + 4i − 6 + 2i = 0 (d) −10 + 4i − 6 + 2i = 0

• 3.7 In the circuit of Fig. 3.49, current i1 is:

  • (a) 4 A (b) 3 A (c) 2 A (d) 1 A