Problem 6: Consider the circuit in the following diagram, where the resistances
ID: 1657911 • Letter: P
Question
Problem 6: Consider the circuit in the following diagram, where the resistances are R1 = 0.45R, R2-2R, R3-0.55R, and R,-6R, where R = 14 . The circuit is connected to a V = 4.9 V source. R2 R4 ©theexpertta.com Part (a) Input an expression for the equivalent resistance of the circuit in terms of R1, R2, R3, and R4. Expression eq Select from the variables below to write your expression. Note that all variables may not be required , , , a, b, d, g, h, j, k, m, P, Ri, R2,R3, R4, Req, S, t, V Part (b) Input an expression using the Req and V for power expended by the circuit. Expression: Select from the variables below to write your expression. Note that all variables may not be required , , , d, g, h, m, P, Ri,R2, R3, R4, Req, t, v Part (c) What is the power P, in watts? Numeric : A numeric value is expected and not an expression.Explanation / Answer
(a)
R1,R2& R3 are in parallel , hence their equivalent resistance would be :
R=1/( 1/R1 + 1/R2 + 1/R3 )
And R4 is connected in series with this resistance.
So total equivalent resistance
Req=[1/ (1/R1+1/R2+1/R3)]+R4
= [R1R2R3/(R2R3+R1R3+R1R2)] + R4
= [ R1R2R3+R2R3R4+R1R3R4+R1R2R4]/(R2R3+R1R3+R1R2)
(b)
Power (P)= I2Req or V2/Req
(C)
Req= [1/{(1/6.3)+(1/28)+(1/7.7)}] +84
= 87.08 ohm
P= V2/Req= (4.9)2/87.08= 0.275 watt