PHYSICS MIDDLE SCHOOL

Please Help Meee.. Will give brainlest What is the total resistance in the circuit depicted? Use R1 = 7.00 Ω, R2 = 8.00 Ω, R3 = 4.00 Ω, R4 = 6.00 Ω, R5 = 3.00 Ω, R6 = 8.00 Ω.

36.0 Ω

19.6 Ω

18.3 Ω

25.9 Ω
malec59 avatar

Answers

Answer 1
Answer:

Answer:

Total resistance is 19.6 Ω for the circuit .

Explanation:

Given:

Six resistors of a circuit namely R_1 to R_6 .

Where

R_1 and R_2 are in series.

R_3 and R_4 are in parallel.

R_5 and R_6 are in parallel.

We know that continuous resistor which are in series are added up to find the equivalent resistance.

Similarly resistors which are arranged in parallel their equivalent resistance is (R_3R_5)/(R_3+R_5)  for the case of R_3 and R_5.

According to the question:

Total resistance = R(equivalent) :

⇒ R_(eq_) = R_1+R_2+((R_3R_4)/(R_3+R_4) )+((R_5R_6)/(R_5+R_6) )

⇒ R_(eq_)=7+8+((4* 6)/(4+6) )+((3* 8)/(3+8) )

⇒ R_(eq_)=7+8+((24)/(10) )+((24)/(11) )

⇒ R_(eq_)=15+(2.4 )+(2.18)

⇒  R_(eq_)=19.58 Ω

So the total resistance of the circuit depicted is 19.58 Ω approximated to nearest tenth that is 19.6 Ω

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