Two 8.7 \\Omega resistors are connected in parallel, as are two 5.0 \\Omega resi

Question

Two 8.7 Omega resistors are connected in parallel, as are two 5.0 Omega resistors. These two combinations are then connected in series in a circuit with a 16 V battery.

Explanation / Answer

Lets say R1 and R2 are 8.7 ohm resistors and R3 and R4 are 5.0 ohm resistors first we will will imagine there are two resistors are in series which i mean R1 combined with R2 forms one resistors and R3 combines with R4 to form the second resistor.And then we can know how much voltage drops in each newly formed resistors. so R12 = (R1*R2)/(R1+R2) => (8.7*8.7)/(17.4) = 4.35 ohm R34 = (R3*R4)/(R3+R4) => (5.0*5.0)/(10) = 2.5 ohm R equivalent = R12 + R34 = 4.35+2.5 = 6.85 OHM now the voltage of battery is 16V and by ohm's law V=IR I = vb/Req => (16 ) / (6.85) = 2.335766423 A since R12 and R34 are in series the current through each resistor is 2.3358 A [by each i mean R12 and R34 not R1,R2,R3,R4] we can now know the voltage drop in each resistor.lets assume v1 is voltage drop across R12 and v2 is voltage drop for R34. =>v1 = (I)( R12) = (2.3358)*(4.35) = 10.16 v V2 = (I)(R34) = (2.3358)*(2.5) = 5.84 v now because R1 and R2 is in parallel the voltage across R1 and R2 is same ,meaning 10.16v .this applies same to R3 and R4 meaning 5.84 V now we can easily find the current in each resistor I1 = V1/R1 = 10.16/8.7 = 1.168 A I2 = V1/R2 = 10.16/8.7 = 1.168 A I3 = V2/R3 = 5.84/5.0 =1.168 A I4 = V2/R4 = 5.84/5.0 =1.168A I went the long way because in future you will get resistors with different values and more complex combinations and this is the whole process of solving these kinds of questions.

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