Question #168231

Given: Resistor 1= 5 ohm, Resistor 2= 12 ohm, resistor 3= 6 ohm, resistor 4= 6 ohm and resistor 5= 30 ohm.

1. Suppose the four resistors are connected in parallel. What is the resistance of the combined resistors?

2. Suppose the four resistors are connected in parallel. Find (a) equivalent resistance (b) current flowing through each resistor (c) voltage across each resistor and (d) power dissipated by each resistor.

3. Find the equivalent resistance of the combination of resistance shown.





1
Expert's answer
2021-03-04T17:23:21-0500

Given resistance are:-

R1=5Ω,R2=12Ω,R3=6Ω,R4=6Ω,R5=30ΩR_1=5\Omega, R_2=12\Omega,R_3=6\Omega, R_4=6\Omega,R_5=30\Omega


(i) As The resistors are connected in Parallel

1Reqv.=1R1+1R2+1R3+1R4+1R5\dfrac{1}{R_{eqv.}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5}


=15+112+16+16+130=\dfrac{1}{5}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{30}


=12+5+10+10+260=3960=\dfrac{12+5+10+10+2}{60}=\dfrac{39}{60}


Reqv.=6039=1.538ΩR_{eqv.}=\dfrac{60}{39}=1.538\Omega


(ii) If The four resistor are connected in parallel then


1Reqv.=1R1+1R2+1R3+1R4\dfrac{1}{R_{eqv.}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}


=15+112+16+16=\dfrac{1}{5}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{6}


=12+5+10+1060=3760=\dfrac{12+5+10+10}{60}=\dfrac{37}{60}


Reqv.=6037=1.62ΩR_{eqv.}=\dfrac{60}{37}=1.62\Omega


(a) Let the voltage across the circuit is V, As all the resistance are connected in Parallel

So Volteage Across each resistor is V.


(b) Current in R1=V5AR_1=\dfrac{V}{5} A

Current in R2=V12AR_2=\dfrac{V}{12} A


Current in R3=V6AR_3=\dfrac{V}{6} A


Current in R4=V6AR_4=\dfrac{V}{6} A

Current in R5=V30AR_5=\dfrac{V}{30} A


(c) Power disisipiaton in R1=V25R_1=\dfrac{V^2}{5} Watt


Power disisipiaton in R2=V212R_2=\dfrac{V^2}{12} Watt


Power disisipiaton in R3=V26R_3=\dfrac{V^2}{6} Watt


Power disisipiaton in R4=V26R_4=\dfrac{V^2}{6} Watt


Power disisipiaton in R5=V230R_5=\dfrac{V^2}{30} Watt



(iii) Equivalent resistance is same as the combined i.e. 1.538Ω1.538\Omega


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