Two resistors, 44.8 and 64.0, are connected in parallel. The current through the

Question

Two resistors, 44.8 and 64.0, are connected in parallel. The current through the 64.0- resistor is 2.95A. (a) Determine the current in the other resistor. (b) What is the total power supplied to the two resistors?

Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.01 A. When the resistors are connected in parallel to the battery, the total current from the battery is 9.46 A. Determine the two resistances.

Explanation / Answer

Quetion 1

Part a

when resistors are connected in parallel
the potential difference across their ends will be the same.
hence 44.8*2.95=64.0*i
so i=(44.8*2.95)/(64.0)

i = 2.065 A

Part b

power=V*i
v=iR
so power=i*iR
so the totalpower=2.065*2.065*64.0+2.95*2.95*44.8=272.91+389.87=662.78 W

Question 2

when resistances are connected in series, current = 1.01 A

=> 12/(R1+R2) = 1.01

=> (R1+R2)=12/1.01

=> R1+R2 = 11.881 ohm -----------(1)

when R1 and R2 are in parallel, current = 11.7 amps

=> 12/[(R1R2)/(R1+R2)] = 9.46, but (R1+R2)= 11.881

=> 12/[(R1R2)/(11.881)] = 9.46

=> R1R2=(12*11.881)/(9.46)

=> R1R2 = 15.071 ohm^2 -------(2)

(R1- R2) = sqrt[(R1+R2)^2 -4R1R2]

(R1- R2) = sqrt[(11.881)^2 -4*(15.071)]

=> R1- R2 = 8.993 ohms ---------(3)

from (1) and (3),

R1 = 10.437 ohm

R2 = 11.881 - 10.437 = 1.444 ohm

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