resistances are R1 = 2.2 ?, R2 = 4.4 ? and R3 = 6.6 ?, respectively, and a new v

Question

resistances are R1 = 2.2 ?, R2 = 4.4 ? and R3 = 6.6 ?, respectively, and a new voltage source is provided. If the current measured in the 6.6-? resistor is 2.5 A, find the following. (a) the potential difference provided by the new battery and the currents in each of the remaining resistors e m f . What factors affect the potential difference across a radiator? V I1 = A (I1 similar to figure above) I2 = A (I2 similar to figure above) (b) the power delivered to each resistor and the total power P1 = W P2 = W P3 = W Ptot = W (c) the equivalent resistance Req = ? (d) the total current and the power dissipated by the equivalent resistor I = A Peq = W

Explanation / Answer

here,

R1 = 2.2 ohm

R2 = 4.4 ohm

R3 = 6.6 ohm

if current in 6.6 ohm , I3 = 2.5 A

as the resistances are in series

potential differnce across resistors is same

I1*R1 = I2 * R2 = I3 * R3

I1 * 2.2 = I2 * 4.4 = 2.5 * 6.6

then ,

I1 = 7.5 A , I2 = 3.75 A

(a)

the potential diffrence provided by new battery , V = ( I1 * R1)

V = 16.5 V

I1 = 7.5 A , I2 = 3.75 A

I3 = 2.5 A

(b)

the power , P1 = I1^2 * R1

P1 = 123.75 W

P2 = I2^2 * R2 = 61.88 W

P3 = I3^2 * R3 = 41.25 W

Ptot = V * (I1 + I2 + I3)

Ptot = 226.88 W

(c)

the equivalent resistance , R = V/( I1 + I2 + I3)

R = 1.2 ohm

(d)

I = ( I1 + I2 + I3) = 13.75 A

Peq = 226.88 W

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