CircuitsCircuits

Which of the equations is valid for the circuit shown below? Which of the equations is valid for the circuit shown below?

A) 2 – I2 – I11 – 2I – 2I22 = 0 = 0

B) 2 – 2I2 – 2I11 – 2I – 2I2 2 – 4I– 4I3 3 = 0= 0

C) 2 – I2 – I11 – 4 – 2I – 4 – 2I2 2 = 0= 0

D) II33 – 2I – 2I2 2 – 4I– 4I2 2 = 0= 0

E) 2I2I11 – I – I11 – 4I – 4I2 2 = 0= 0

2 V

2

2 V 6 V

4 V

3 1

1

I1 I3

I2

Suppose x + y – 5 = 2 and 3x – 2y + 1 = 0. What is x?

A) zero

B) 13 / 5

C) 2 / 3

D) 7

E) 17 / 4

Solve first equation for y:

x + y = 7

y = 7 – x

Plug into second equation

3x – 2(7 – x) + 1 = 0

3x – 14 + 2x + 1 = 0

5x – 13 = 0

5x = 13

x = 13 / 5

Algebra TestAlgebra Test

In the circuit below, if all the resistors are identical, what is the voltage drop across resistor R2?

= 8 V

R1

R2

V = –8 VLoop rule

(bottom left loop)

V

V Same current as R1 and same resistance V =V

V = ?

V

right loopV V = 0

How do the currentscompare here and here?

Give me an expression that describes the current I1 through R1

In the circuit below all the resistors are identical.

= 8 V

R1

R2

V = –8 VLoop rule

(bottom left loop)

V

V Same current as R1 and same resistance V =V

V = ?

V

right loopV V = 0

Both same current, I2

I1 = 2I2

What equation best describes the CW full outermost loop?

1) 8v + I2R1 + I2R I2R2 = 02) -8v + I2R1 + I2R + I2R2 = 03) -8v + 2I2R1 + 2I2R + I2R2 = 04) 8v 2I2R1 2I2R I2R2 = 0

In the circuit below all the resistors are identical.

= 8 V

R1

R2

V = –8 VLoop rule

(bottom left loop)

V

V Same current as R1 and same resistance V =V

V = ?

V

right loopV V = 0

Both same current, I2

I1 = 2I2

What is the voltage drop across resistor R2?

1) 0.2 volts 2) 1.0 volts3) 1.2 volts 4) 1.6 volts5) 2.0 volts 6) 4.0 volts

Determine the magnitudes and Determine the magnitudes and directions of the currents directions of the currents through Rthrough R11 in the figure at right. in the figure at right.

V2 = 6 V

R2 = 15

R1 = 25 V1 = 9 V

Label CurrentsI1

I3

I2

Junction Rule

Loop Rule (1)

Loop Rule (2)

I1 = I2 + I3

+V2 – I1R2 = 0

+I1R2 + I2R1 + V1 = 0

Algebra

1

Sample ProblemSample Problem

Solve Loop Rule (1) for I1

Plug into Loop Rule (2) I2 = –V1 – V2

R1

2

Comments I2 = –3/5 Amp What does the minus sign mean?Current goes in opposite direction of arrow