A) for each arrangement give the algebraic expression for the equivalent resista

Question

A) for each arrangement give the algebraic expression for the equivalent resistance of the diagramed arrangement and evaluate the expression for R=840ohms. B) evaluate the current through each resistor when the combination is connected to a power supply giving 1200V. C) evaluate the power dissipated in each resistor of each combination, and the total power dissipated by each full combination, when the combination is connected to the same 1200V power supply. B) evaluate the current through each resistor when the combination is connected to a power supply giving 1200V. C) evaluate the power dissipated in each resistor of each combination, and the total power dissipated by each full combination, when the combination is connected to the same 1200V power supply. Problem 7

Explanation / Answer

(A)

(a):

R(net) = R + R + R + R = 4R ohm

(b):

R(net) = 1 / ( (1/R)+ (1/R)+ (1/R)+ (1/R) ) = R/4 ohm

(c) :

R(net) = 1 / ( (1/R)+ (1/R) ) + 1 / ( (1/R)+ (1/R) ) = R/2 + R/2 = R ohm

(d) :

R(net) : 1 / ( (1/(R+R))+ (1/(R+R)) ) = 2R/2 = R ohm

(e) :

R(net) = 1 / ( (1/(R+R+R))+ (1/R) ) = (3/4) R ohm

(f):

R(net) = 1 / ( (1/R)+ (1/R)+ (1/R) ) + R = R/3 + R = (4/3) R ohm

(B) :

if V it the across the ends then,

(a) :

Current through each resistance is I = (V) / (Rnet) = (V) / (4R) A

(b):

Current through each resistance is I =  (V) / (Rnet) = (V) / (R/4) = 4V / R A

(c) :

Current through each resistance is I = ( (V) / (Rnet) ) * (1/2) = ( (V) / (R) )* (1/2) = V / (2R) A

(d):

Current through each resistance is I = ( (V) / (Rnet) ) * (1/2) = ( (V) / (R) ) * (1/2) A = V / (2R) A

(e):

Current through each resistance isof upper branch is I = ( (V) / (Rnet) ) * ( (R) / (R + 3R) )=( (V) / ((3/4)R) ) * ( (R) / (R + 3R) )

I = V / (3R) A

Current through each resistance isof upper branch is I = ( (V) / (Rnet) ) * ( (3R) / (R + 3R) )=( (V) / ((3/4)R) ) * ( (R) / (R + 3R) )

I = V / (R) A

(f) :

Current through each resistance of first three R in parallel combination is I = ( (V) / (Rnet) ) * ( 1/3 )=( (V) / ((4/3)R) ) * ( 1/3 )

I = V / (4R) A

Current through resistance in series with set of three parallel combination is I = ( (V) / (Rnet) ) =( (V) / ((4/3)R) )

I = (3V) / (4R) A

(C) :

(a): Power dessipated through each R is P = I^2 * Rnet = (V^2) / (16R) W

Power dessipated through combination is P = V^2 / Rnet = (V^2) / (4R) W

(b)

Power dessipated through each R is P = I^2 * R = (16*V^2) / (R) W

Power dessipated through combination is P = V^2 / Rnet = (4V^2) / (R) W

(c):

Power dessipated through each R is P = I^2 * R = (V^2) / (4R) W

Power dessipated through combination is P = V^2 / Rnet = (V^2) / (R) W

(d) :

Power dessipated through each R is P = I^2 * R = (V^2) / (4R) W

Power dessipated through combination is P = V^2 / Rnet = (V^2) / (R) W

(e):

Power dessipated through each R of upper branch is P = I^2 * R = (V^2) / (9R) W

Power dessipated through each R of lower branch is P = I^2 * R = (V^2) / (R) W

Power dessipated through combination is P = V^2 / Rnet = (4V^2) / (3R) W

(f):

Power dessipated through each R in the first three R in parallel combination is P = I^2 * R = (V^2) / (16R) W

Power dessipated through R in series with the set of three R in parallel combination is is P = I^2 * R

P =  = (9V^2) / (16R) W

Power dessipated through combination is P = V^2 / Rnet = (3V^2) / (4R) W

now put R = 840 ohm and V = 1200 V to get your answer.

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