the equivalent resistance between points A and B in the circuit shown below is 2

Question

the equivalent resistance between points A and B in the circuit shown below is 26 ohm, a) find value of resistance R. b) a 12 v battery is connected to terminals A and B. find the current in each resistors and power dissipated by each resistors University Physics I (PHYS 23261 University Physics II (PHYS 2326) Houston Community College 7/26/2018 Quiz: Chap. 27 & 28 Show your solution to get full credit. 1· The equivalent resistance between points A and B in the circuit shown below is 26 C. (a) Find the value of stae . (b) A 12-V battery is connected to terminals A and B. Find the current in each resistor and the power dissipated by each resistor 5SJ2 and are an pavallel, 55 ? and R ave in sers, S SSAR 2. Find the magnitude and direction of the current (clockwise or counterclockwise) in the figure below is.ov.t ll-SVz?(getc. 2@tw) ? ? 622? 29.89 15.1 ? and the direction 4 current is counter clerkusele.

Explanation / Answer

Given

Circuit consisting of resistances R1 = 12 ohm R2 = 55 ohm, R = ?

R2 = 55 ohm and R3 in parallel combination  

and the resusltant resistance is R = 26 ohm

the total resistance is26 = 12+((55*R3)/(55+R3))

solving for R3 , R3 = 18.780 ohm

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applying the kirchhoff's voltage rule  

as total resistance is 26 ohm and V = 12 V

so the current in the circuit is I = v/r = 12/26 A = 0.46154 A

the potential drop across R1 is v1 = I1*R1 = 0.46154*12 V = 5.53848 V

remaining voltage (12-5.53848 =6.46152 ohm) will be divided across R2,R3

so the current across R2 is I2 = V2/R2 = 6.46152/55 A = 0.1175 A

the current across R3 is I3 = V3/R3 = 6.46152/18.78 A = 0.3441 A

the power dissipated is P = I^2*R

P1 = I1^2*R1 = 0.46154 ^2*12 W = 2.55623 W

throrugh R2 is  

P2 = I2^2*R2 = 0.1175 ^2*55 W = 0.75934 W

throrugh R2 is  

P3 = I3^2*R3 = 0.3441 ^2*18.78 W = 2.22364 W

the answers are  

R = R3 = 18.78 ohm

I1 = 0.46154 A, I2 = 0.1175 a, I3 = 0.3441 A

power dissipation  

P1 = 2.55623 W, P2 = 0.75934W , P3 = 2.22364 W

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Applying kirdhhoff's voltage rule  

clockwise direction  

15- 8.50*I +11.5-6.22*I-15.1*I = 0

solving for I , I = 0.88867 A = 0.89 A in the clock wise dierection

we know the potential will be positive if we move from -ve terminal of the battery to the +ve terminal and it is the direction of the curerrent in the circuit

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