The Physics of Everyday Phenomena
The Physics of Everyday Phenomena
8th Edition
ISBN: 9780073513904
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
Question
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Chapter 13, Problem 1SP

(a)

To determine

The equivalent resistance of the circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 1SP

The equivalent resistance is 4Ω

Explanation of Solution

Given info: The resistances are 6Ω and 12Ω

Write the equation to find the equivalent resistance.

Req=R1R2R1+R2

Here,

Req is the equivalent resistance

R1 is the first resistance

R2 is the second resistance

Substitute 6Ω R1 and 12Ω for R2 to get Req

Req=6Ω×12Ω6Ω+12Ω=4Ω

Conclusion:

Therefore, the equivalent resistance is 4Ω

(b)

To determine

The total current flowing through the circuit.

(b)

Expert Solution
Check Mark

Answer to Problem 1SP

The total current is 0.125A

Explanation of Solution

Here the equivalent resistance of R1 and R2 is in series with R3

Write the equation to find the equivalent resistance of the circuit.

R=Req+R3

Here,

R is the equivalent resistance of the whole circuit

Req is the equivalent resistance of parallel resistors

R3 is the third resistance

Substitute 8Ω for R3 and 4Ω for Req to get R

R=8Ω+4Ω=12Ω

Write the equation to find the current.

I=VR

Here,

I is the current

V is the voltage

R is the resistance

Substitute 1.5V for V and 12Ω for R to get I

I=1.5V12Ω=0.125A

Conclusion:

Therefore, the current is 0.125A

(c)

To determine

Current flowing through 6Ω resistor.

(c)

Expert Solution
Check Mark

Answer to Problem 1SP

Current is 0.0833A

Explanation of Solution

Write the equation to find current.

I=VR

Here,

I is the current

V is the voltage

R is the resistance

Substitute 1.5V for V and 6Ω for R to get I

I=1.5V6Ω=0.0833A

Conclusion:

Therefore, the current flowing is 0.0833A

(d)

To determine

Power dissipated in 8Ω resistor.

(d)

Expert Solution
Check Mark

Answer to Problem 1SP

Power dissipated is 125mW

Explanation of Solution

Write the equation to find the current through 8Ω resistor

I=VR

Here,

I is the current

V is the voltage

R is the resistance

Substitute 1.5V for V and 8Ω for R to get I

I=1.5V8Ω=0.1875A

Write the equation to find the power.

P=I2R

Here,

P is the power

I is the current

R is  the resistance

Substitute 0.1875A for I and 8Ω for R to get P

P=(0.1875A)2×8Ω=125mW

Conclusion:

Therefore, the power consumed is 125mW

(e)

To determine

Is the current through eight ohm resistor greater than six ohm resistor?

(e)

Expert Solution
Check Mark

Answer to Problem 1SP

Yes, it is greater.

Explanation of Solution

The current through eight ohm resistor is greater than the current flowing through six ohm resistor since the eight ohm resistor is connected serially and the six ohm resistor is connected parallel.

Conclusion:

Yes it is greater.

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The Physics of Everyday Phenomena

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