eplain in detail how current is solved in part B and part C please list all rule

Question

eplain in detail how current is solved in part B and part C

please list all rules and show each step of the picture

answers: a.) 196 ohms b.) 0.0305 A c.) 2.18 V d.) 0.0467 show all steps and equations! The circuit shown in the figure below has a 6.00-V battery. The resistors have resistances of R1 # 50.0 ?, R2-250.0 ?, R3-100.0 ?, R4 2. 30.0 ?, and Rs-45.0 ?. a) What is the equivalent resistance for the circuit? RI 83 84 b) What is the current through resistor 4? c) What is the potential difference across resistor 3?

Explanation / Answer

2.

(a)

Here in the circuit , R2 and R3 are in parallel and their equivalent resistance is in series with R1, R4 and R5

Equivalent resistance can be given as :

Req = R1 + [(R2*R3)/(R2+R3)] + R4 + R5

= 50 + [(250*100)/(250+100)]+30+45

= 196.43 ohm

(b)

Since we know that, current through resistors in the series combination is same through each component.

So current through resistor 4 can be given as :

I = V/Req = 6/196.43= 0.0305 A

(c)

Potential difference across resistor R3 :

As we know that , potential difference across each component in parallel is same.

So , at resistor R3 : V = I *[R2R3/(R2+R3)]

V = (0.0305 A)* [(250*100)/350]

= 2.18 V

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