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Question

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R1=7.0 ohms R2=4.0 ohms R3= 7.0 ohms R4= 8.0 ohms R5= 8.0 ohms R6=5.0 ohms and the voltage is 15.0V.
What is the voltage across R6? What is the voltage across the combination of R1, R2, R3, R4, and R5? What is the voltage across R1? What is the voltage across the combination of R2, R3, R4, and R5? What is the current going through R1? What is the current going through the combination of resistors R2, R3, R4, and R5? What is the current going through R5? What is the current going through the combination of resistors R2, R3, and R4? What is the voltage across R5?What is the voltage across the combination of R2, R3, and R4? What is the voltage across R2? What is the voltage across R3? What is the voltage across R4? What is the current going through R2? What is the current going through R3? What is the current going through R4?

Explanation / Answer

Here is a redrawing of the entire circuit diagram, yes this is a correct redrawing.

The first thing is to find the total amount of current in the circuit, by combining all the resistors.

First lets go with the three parallel resistors

1/R234= 1/R2+1/R3+1/R4

1/R234= 1/4+1/7+1/8

R234= (29/56)^-1= 56/29

Now combine the parallel resistors with R5

R2345= 56/29+8= 288/29

Now combine the big resistor with R_1 that is in parallel

1/R12345= 1/7+ 29/288

R12345= (491/2016)^-1= 2016/491

Now combine R_6 to get the total resistance

R123456= 2016/491+5= 4471/491

Use Ohm's Law to find the total current going through the circuit

= IR123456

I= /R123456= 15V/ (4471/491 )= 1.64728 A

Now it is possible to answer the other questions

For the Voltage across R_6, use Ohm's law again

V_6= I*R_6= (1.64728A)*5, V_6= 8.236 V

In finding the Use Ohm's Law except use R_12345 for the voltage across the rest of resistors

V_12345= I*R_12345= (1.64728 A)*(2016/491)= 6.764 V

When Resistors are in parallel, the voltage is going to be the same across all branches in parallel

So the Voltage across R_1 is going to be the same calculate for the combination of Resistors

R_1,V_1= 6.764V

For the combination of R_2,R_3,R_4,R_5 will be the same amount because it is in parallel with R_1

R_2345, V_2345 6.764 V

To find the Current in R_1 Use Ohm's Law again

V_1=I_1*R_1, I_1= V_1/R_1= (6.764V)/7 = .966 A

To find the current across the combination of resistors, you can either subtract (Kirchoff's Junction Rule)it from the total current or use Ohm's Law again and use R_2345

1.64728- .966 A= .681 A

V_2345= I_2345*R_2345, I_2345= V_2345/R_2345= (6.764V)/(288/29)= .681 A

Since we have the amount of current going through the upper branch it is going to be same amount of current going through R_5, and R_234

I_5= I_2345= .681 A

I_234= I_2345= .681 A

To find the voltage across R_5 Use Ohm's Law

V_5= I_5 * R_5= (.681A) (8)= 5.448 V

Use Ohm's Law again for the Voltage across the combination of resistors (R_2,R_3, R_4)

V_234= I_234*R_234= (.681A)(56/29)= 1.315 V

Now for the last branch of resistors that are in parallel, so the will have the same amount of voltage across them

V_2= V_3=V_4=V_234= 1.315V

Now Use Ohm's Law to find the amount of current through them

V_2= I_2*R_2, I_2= V_2/R_2=(1.315/4)= .329 A

V_3= I_3*R_3, I_3= V_3/R_3=(1.315/7)= .188 A

V_4= I_4*R_4, I_4= V_4/R_4= (1.315/8)= .164A

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