Scribd
Upload
124 views3 pages

Electrical Circuit Solutions

This document summarizes an exam for an electric circuit analysis course. It includes 5 practice problems and their solutions. Problem 1 asks students to find the current from a given charge function. Problem 2 asks students to sketch power from given current and voltage graphs and calculate total energy. Problem 3 asks for ammeter and voltmeter readings in a circuit. Problem 4 asks for a voltmeter reading. Problem 5 asks for current through a resistor and voltage across a current source.

Uploaded by

Pao JJ
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
124 views3 pages

Electrical Circuit Solutions

This document summarizes an exam for an electric circuit analysis course. It includes 5 practice problems and their solutions. Problem 1 asks students to find the current from a given charge function. Problem 2 asks students to sketch power from given current and voltage graphs and calculate total energy. Problem 3 asks for ammeter and voltmeter readings in a circuit. Problem 4 asks for a voltmeter reading. Problem 5 asks for current through a resistor and voltage across a current source.

Uploaded by

Pao JJ
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 3

ECE 2004: Electric Circuit Analysis

Fall 2006, 3 credits, CRN: 91777, 91778, 91779, 91780, 96346: Test#1
Dr. Pushkin Kachroo
The Bradley Department of Electrical and Computer Engineering, Virginia Tech, Blacksburg, VA 24061-0111,
pushkin@vt.edu
PROBLEM 1: The total charge q(t), in Coulombs, that enters the terminal of an element is
⎧ 0 t<0
⎪
q (t ) = ⎨ 2t 0≤t ≤2
⎪3 + e − 2 ( t − 2 ) t>2
⎩
Find the current i(t) and sketch its waveform for t ≥ 0 . (10 points)
SOLUTION 1:

⎧0 t <0
dq ( t ) ⎪
i (t ) = i (t ) = ⎨2 0< t < 2
dt ⎪ −2( t − 2 )
⎩−2e t >2

PROBLEM 2: The current through and voltage across an element vary with time as shown in figure. Sketch
the power delivered to the element for t ≥ 0 . What is the total energy delivered to the element between t = 0
to t = 25 s? (10 points)

SOLUTION 2:
30
for 0 ≤ t ≤ 10 s: v = 30 V and i = t = 2t A ∴ P = 30(2t ) = 60t W
15
25
for 10 ≤ t ≤ 15 s: v ( t ) = − t + b ⇒ v (10 ) = 30 V ⇒ b = 80 V
5
v(t ) = −5t + 80 and i (t ) = 2t A ⇒ P = ( 2t )( −5t +80 ) = −10t 2 +160t W
30
for 15 ≤ t ≤ 25 s: v = 5 V and i (t ) = − t +b A
10
i (25) = 0 ⇒ b = 75 ⇒ i (t ) = −3t + 75 A
∴ P = ( 5 )( −3t + 75 ) = −15t + 375 W
Energy = ∫ P dt = ∫ 60t dt + ∫10 (160t −10t 2 ) dt + ∫15 ( 375−15t ) dt
10 15 25

10 15 25
= 30t 2 + 80t 2 − 10 t 3 + 375t − 15 t 2 = 5833.3 J
0 3 10 2 15

PROBLEM 3: Find the ammeter and the voltmeter readings in figure below (10 points)

SOLUTION 3:
i = −2 Amps ;
Using KVL, we get 12 − 4i − v = 0
v = 12 − 4i = 12 − 4(−2) = 20 Volts
PROBLEM 4: Find the voltmeter reading in figure below (10 points)

SOLUTION 4:
We can label the circuit as shown.

The subscripts suggest a numbering of the


circuit elements. Apply KCL at the top node of
the current source to get

i1 = i 2 + 0.25

Apply Ohm’s law to the resistors to get

v1 = 20 i1 and v 2 = 60 i 2 = 60 ( i1 − 0.25 ) = 60 i1 − 15

Apply KVL to the outside to get


v 2 + 80 i1 + v1 = 0 ⇒ ( 60 i 1 − 15 ) + 80 i1 + 20 i1 = 0 ⇒ i1 =
15
160
= 0.09375 A

Finally,
v m = 80 i1 = 80 ( 0.09375 ) = 7.5 V
PROBLEM 5: Find the current flowing through the 2 Ohms resistor (on the left) in the direction of the arrow,
and the voltage across the 3A current source. (10 points)

SOLUTION 5:
Let us call the current flowing as shown by the arrow to be i , and the voltage across the 3 Amps current
source (that means voltage at the top minus the voltage at the bottom of the current source) to be v.

Applying KVL gives:


3 − 2i − 8 − v = 0 (1)
Note that the current flowing downwards through the 2 Ohm resistor on the right (given by Ohm’s law) is v/2.

Applying KCL st the top node gives


v
i + 3 = + 1.25 (2)
2
Solving (1) and (2) gives
i = −2.125 Amps; v = −0.75 Volts

You might also like