Practice Problems (KEY)

Practice Problems (KEY)#

Problem 1

  • Nodes: 4   Nodes: 6

  • Loops: 3   Loops: 7

  • Nodes: 4   Nodes: 4

  • Loops: 1   Loops: 7

Problem 2

\[V_1 = (2\,\text{mA})(1.5\,\text{k}\Omega) = 3\,\text{V}\]
\[V_S - V_1 - V_R = 0\]
\[V_R = V_S - V_1 = 9\,\text{V} - 3\,\text{V} = \boxed{6\,\text{V}}\]
\[V_R - V_P - V_{\text{out}} = 0\]
\[V_P = V_R - V_{\text{out}} = 6\,\text{V} - 2\,\text{V} = \boxed{4\,\text{V}}\]
Problem 3

Given \(i_2 = -2\,\text{A}\), \(i_3 = -5\,\text{A}\), and \(i_4 = 4\,\text{A}\):

\[i_5 + i_4 - i_3 = 0 \implies i_5 = -i_4 + i_3 = -(4\,\text{A}) + (-5\,\text{A}) = \boxed{-9\,\text{A}}\]
\[i_1 + i_2 + i_3 - i_4 = 0 \implies i_1 = -i_2 - i_3 + i_4 = -(-2) - (-5) + 4 = \boxed{11\,\text{A}}\]
Problem 4

(a) Nodes: 3   Loops: 3

(b) KVL and KCL equations:

  • \(V_Q - V_P - V_R = 0\)

  • \(V_R - V_S = 0\)

  • \(V_Q - V_P - V_S = 0\)

  • \(I_Q + I_P = 0\)

  • \(-I_P - I_R - I_S = 0\)

  • \(-I_Q + I_R + I_S = 0\)

(c) Power for each component (\(Q\) is the source, so it supplies the \(20\,\text{W}\) that the other three consume):

  • \(V_S = 4\,\text{V}\), \(V_R = 4\,\text{V}\), \(V_P = 6\,\text{V}\), \(V_Q = 10\,\text{V}\)

  • \(I_S = 1\,\text{A}\), \(I_R = 1\,\text{A}\), \(I_P = -2\,\text{A}\), \(I_Q = 2\,\text{A}\)

  • \(P_S = 4\,\text{W}\), \(P_R = 4\,\text{W}\), \(P_P = 12\,\text{W}\), \(P_Q = 20\,\text{W}\)

Problem 5

KVL Equations

KCL Equations

\(-V_G - 2\,\text{V} + V_J = 0\)

\(4\,\text{A} + I_H - 3\,\text{A} = 0\)

\(2\,\text{V} - 7\,\text{V} - V_K = 0\)

\(3\,\text{A} + I_J - 5\,\text{A} - 6\,\text{A} = 0\)

\(-V_L + V_K + 17\,\text{V} = 0\)

\(I_M + I_N + 5\,\text{A} = 0\)

\(-V_J + V_L - 6\,\text{V} = 0\)

\(6\,\text{A} - I_N - I_H = 0\)

\(V_G = 4\,\text{V}\)

\(I_M = -I_N - 5\,\text{A}\)

Solving the KVL equations in order: \(V_K = 2 - 7 = -5\,\text{V}\), \(V_L = V_K + 17 = 12\,\text{V}\), \(V_J = V_L - 6 = 6\,\text{V}\), and \(V_G = V_J - 2 = 4\,\text{V}\).

Solving the KCL equations in order: \(I_H = 3 - 4 = -1\,\text{A}\), \(I_J = 5 + 6 - 3 = 8\,\text{A}\), \(I_N = 6 - I_H = 7\,\text{A}\), and \(I_M = -7 - 5 = -12\,\text{A}\).

\[\boxed{V_G = 4\,\text{V},\ V_J = 6\,\text{V},\ V_K = -5\,\text{V},\ V_L = 12\,\text{V}} \qquad \boxed{I_H = -1\,\text{A},\ I_J = 8\,\text{A},\ I_M = -12\,\text{A},\ I_N = 7\,\text{A}}\]
Problem 6

A satellite control module is connected to a data processing element modeled as a single \(12.5\,\text{k}\Omega\) resistor (\(R_2\)). The system is powered by a \(5\,\text{V}\) solar cell (\(V_S\)). Control module options:

  • \(R_1 = 5\,\text{k}\Omega\)

  • \(R_1 = 10\,\text{k}\Omega\)

  • \(R_1 = 15\,\text{k}\Omega\)

The two resistors are in series, so \(I_1 = I_2 = \dfrac{V_S}{R_1 + R_2}\), \(V_1 = I_1R_1\), \(V_2 = I_2R_2\), and each power is \(P = IV\).

Option

Vs

R1

V1

V2

I1

I2

P1

P2

Ps

A

5 V

5 kΩ

1.43 V

3.57 V

285.7 µA

285.7 µA

408.2 µW

1.020 mW

1.429 mW

B

5 V

10 kΩ

2.22 V

2.78 V

222.2 µA

222.2 µA

493.8 µW

617.3 µW

1.111 mW

C

5 V

15 kΩ

2.73 V

2.27 V

181.8 µA

181.8 µA

495.9 µW

413.2 µW

909.1 µW
Problem 7

Requirements:

  • Total current must not exceed \(250\,\mu\text{A}\)

  • Voltage across the control module (\(V_1\)) must be at least \(2\,\text{V}\)

  • Option A: \(285.7\,\mu\text{A} > 250\,\mu\text{A}\) and \(V_1 = 1.43\,\text{V} < 2\,\text{V}\). Fails both requirements.

  • Option B: \(222.2\,\mu\text{A}\) and \(V_1 = 2.22\,\text{V}\). Meets both.

  • Option C: \(181.8\,\mu\text{A}\) and \(V_1 = 2.73\,\text{V}\). Meets both.

Options B and C are viable.