Question

For the circuit shown below, find the magnitude and the direction of the current through each resistor and the power supplied by each battery, using the following values: R1=4.53 ohms, R2=6.63 ohms, R3=8.94 ohms, R4=8.17 ohms, R5=3.78 ohms, R6=17.8 ohms, R7=5.12 ohms, Vemf1=7.22 V, Vemf2=11.6V.

R3 Rs Vernf,l emf,I Rs R. WA emf 2

Answers have to be in the form of this

Please show all work and explain.

Thank you

Answer 1

What my strategy is to do I will take our current from V₂ and distribute it to different branches of the circuit. At each node i will distribute the current. I will try to use as minimum as possible unknown currents so that it will be easy to solve equations containing values of currents as unknowns.

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Sign convention for the voltages are as follows:

1) If we are moving in clockwise direction while applying Kirchoff's voltage law (KVL) in a loop then our currents in our directions will give negative voltage across resistor (ex- if i am moving in the right direction while calculating voltages at each resistor then the value of voltage across that resistor will be negative otherwise positive).

2) Going from lower to higher potential will be taken as positive when we move across battery (if current is entering into negative plate its positive or otherwise its negative) or if current is entering into postive plate voltage across battery will be taken as negative otherwise positive).

Now applying KVL across loop 1 gives

$-(i-i_{1})(R_{3}+R_{1})-i_{2}R_{4}+i_{1}R_{5}=0$

Substituing values of R's, we get equation 1

$(17.25)i_{1}-(13.47)i-(8.17)i_{2}=0$

KVL across loop 2 we

$-(i-i_{1}-i_{2})R_{2}+V_{1}+i_{2}R_{4}=0$

reduces to equation B

$(6.63)i_{1}-(6.63)i+(14.80)i_{2}=-7.22$

Across loop 3

$-i_{1}R_{5}-V_{1}-i(R_{6}+R_{7})=0$

this reduces to equation C

$-(3.78)i_{1}-(22.92)i=7.22$

Similarly from 4th loop we get

$-(i-i_{1})(R_{3}+R_{1})-(i-i_{1}-i_{2})R_{2}-i(R_{6}+R_{7})+V_{2}=0$

which gives equation D

$-(43.02)i+(20.1)i_{1}+6.63i_{2}=-11.6$

From equation C we can find i in terms of i₁ and then substitute in any two equation to make the form in i₁ and i₂.

$i=-(.164)i_{1}-(.315)$

Equation D and B reduces to

$(27.155)i_{1}+(6.63)i_{2}=-13.55$

$(7.71)i_{1}+(14.80)i_{2}=-7.22$

Solving it gives values of i,i1 and i2

(here value negative and positive sign shows the direction of the current)

Current through R3 and R1 = i - i1 = +.185A (+ means we assumed the direction correct so clockwise as assumed)

I through R4 = i2 = -.26 A (counter clockwise)

I through R2 = (i - i1 - i2) = + .445 A clockwise

I through R 5 = i1 = - .43 A ( clockwise)

I through R6 and R7 = i = -.245 A (counter clockwise)

Power through V1 = V1x(i1+i2) = 7.22x(.69) = 4.98 W

Power through V2 = V2xi = 11.6 x .245 = 2.84 W