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For the circuit shown in FIGURE 28.59, find the current through and the potential difference across each resistor. Place your results in a table for ease of reading.

Short Answer

Expert verified

We have given the circuit.

We need to find the current through and the potential difference across each resistor.

Step by step solution

01

Given information

We have given the circuit.

We need to find the current through and the potential difference across each resistor.

02

Simplify

When the resistors are connected at both the ends of the point then it is said that the resistors are connected in parallel. This is obvious when the resistors are aligned side by side. The potential difference across the resistors is the same in parallel connection. Equation (28.24)shows the equivalent resistance for the parallel connection in the form

1Req=1R1+1R2+...+1RN

The two resistors 6ฮฉand 12ฮฉare in parallel, so we use the equation (1)to get their combination by

1R1,eq=16ฮฉ+112ฮฉR1,eq=16ฮฉ+112ฮฉ-1R1,eq=4ฮฉ

The resistor 4ฮฉis in series with a combination R1,eq,so they have the same current. We apply the loop rule to get the current through resistors 4ฮฉand the combination R1,eq. Traveling clockwise in the left loop and get the next

V=0Iโˆˆ-(2ฮฉ)-I4(4ฮฉ)-I4R1,eq=024-2I-4I4-4I4=024-2I-8I4=0 (2)

Appling the loop rule on the right loop and travel clockwise to get

V=0-I8(8ฮฉ)+I4R1,eq+I4(4ฮฉ)=0-8I8+4I4+4I4=0I4=I8

03

Simplify

The current Ifrom the junction rule equals I=I4+I8=2I4=2I8.Use this into equation (2)to get I4by

24-2I-8I4=024-2(2I4)-8I4=0I4=2A

Hence, the current through 8ฮฉis I8=2A

Using Ohm's law to get the voltage across both resistors by

โˆ†V4=I4R4=(2A)(4ฮฉ)=8V

and

โˆ†V8=I8R8=(2A)(8ฮฉ)=16V

The current Iis the same for resistors 2ฮฉ, so the current through it is calculated by

I2=I4+I8=2A+2A=4A

Use Ohm's law to get the voltage across this resistor by

โˆ†V2=I2R2=(4A)(2ฮฉ)=8V

Both resistors 6ฮฉand 12ฮฉhave the same voltage and their combination current is the same for I4, so using Ohm's law, the voltage across both of them is

โˆ†V6=โˆ†V12=I4R1,eq=(2A)(4ฮฉ)=8V

Again using Ohm's law, the current through 6ฮฉand 12ฮฉwill be

I6=โˆ†V6R6=8V6ฮฉ=1.33A

and

I12=โˆ†V12R12=8V12ฮฉ=0.66A

04

Table

The results are summarised in below's table

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