the circuit shown below, resistor Ri has a resistance of 50.0 obms, resistor Rs

Question

the circuit shown below, resistor Ri has a resistance of 50.0 obms, resistor Rs has a (a) what is the equivalent resistance of Ri and Ra onms, and resistor Rs has a resistance of 35.0 ohms. A steady current flows throuh e hve circult (b) What is the equivalent resistance of all the resistors: Rs, Ra, and Rs (e) If the conventional current through R is 0.28 Amperes, then what is the emf of the battery? Hint First find current in R3 in terms of of battery.) (d) What is the power dissipated in each resistor?

Explanation / Answer

given R1= 50 ohm

R2=30 ohm

R3=35 ohm

a) Eq resistance of R1 , R2

as both R1 and R2 are in parallel

r = R1 x R2/(R1+R2)

r = 50x30 /(50+30)

r = 18.75 ohm

b) Eq resistance of R1 , R2 and R3

as R1 and R2 are in parallel and theril eq is in series with R3

So lets say Eq of R1 R2 and R3 is Req

so Req = r + R3

Req = 18.75 + 35

= 53.75 ohm

c) let the current in R3 = i

we know that from ohm's law

i = E/Req

i = E/53.75

given i = 0.28 A

so 0.28 = E/53.75

E = 0.28 x53.75

E = 15.05 V

d) i = current through R3 = 0.28 A

so P3 = i2R3 = (0.28x0.28) X 35 = 2.74 W

as R1 and R2 are in parallel so the current will be disribuit in opposite ration

i1 = [ R2/(R1+R2) ] x i

= 30/80 x 0.28

= 0.105 A

so heat dissipated in R1 = P1 = I12xR1 = (0.105)2 x 50 = 0.55 W

and now

i2 = [ R1/(R1+R2) ] x i

= 50/80 x 0.28

= 0.175 A

so heat dissipated in R2 = P2 = i22xR2 = (0.175)2 x 30 = 0.92 W

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